RD Sharma XI 2020 2021 Volume 1 Solutions for Class 12 Commerce Maths Chapter 13 - Complex Numbers

Share with your friends

RD Sharma XI 2020 2021 Volume 1 Solutions for Class 12 Commerce Maths Chapter 13 Complex Numbers are provided here with simple step-by-step explanations. These solutions for Complex Numbers are extremely popular among class 12 Commerce students for Maths Complex Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2020 2021 Volume 1 Book of class 12 Commerce Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2020 2021 Volume 1 Solutions. All RD Sharma XI 2020 2021 Volume 1 Solutions for class 12 Commerce Maths are prepared by experts and are 100% accurate.

Page No 13.3:

Question 1:

Evaluate the following:
(i) i⁴⁵⁷
(ii) i⁵²⁸
(iii) $\frac{1}{i^{58}}$
(iv) $i^{37} + \frac{1}{i^{67}}$
(v) ${(i^{41} + \frac{1}{i^{257}})}^{9}$
(vi) $(i^{77} + i^{70} + i^{87} + i^{414})^{3}$
(vii) $i^{30} + i^{40} + i^{60}$
(viii) $i^{49} + i^{68} + i^{89} + i^{110}$

Answer:

$(i) i^{457} = i^{4 \times 114 + 1} = {(i^{4})}^{114} \times i = i (∵ i^{4} = 1) (ii) i^{528} = i^{4 \times 132} = {(i^{4})}^{132} = 1 (∵ i^{4} = 1) (iii) \frac{1}{i^{58}} = \frac{1}{i^{4 \times 14 + 2}} = \frac{1}{{(i^{4})}^{14} \times i^{2}} = \frac{1}{i^{2}} (∵ i^{4} = 1) = - 1 (∵ i^{2} = - 1) (iv) i^{37} + \frac{1}{i^{67}} = i^{4 \times 9 + 1} + \frac{1}{i^{4 \times 16 + 3}} = {(i^{4})}^{9} \times i + \frac{1}{{(i^{4})}^{16} \times i^{3}} = i - \frac{1}{i} (∵ i^{3} = - i) = i - \frac{1}{i} \times \frac{i}{i} = i - \frac{i}{i^{2}} = i - (- i) (∵ i^{2} = - 1) = 2 i (v) {(i^{41} + \frac{1}{i^{257}})}^{9} = {(i^{4 \times 10 + 1} + \frac{1}{i^{4 \times 64 + 1}})}^{9} = {[{(i^{4})}^{10} \times i + \frac{1}{{(i^{4})}^{64} \times i}]}^{9} = {(i + \frac{1}{i})}^{9} (∵ i^{4} = 1)$

$= {(i + \frac{i}{i^{2}})}^{9} = {(i - i)}^{9} (∵ i^{2} = - 1) = 0$

$(vi) {(i^{77} + i^{70} + i^{87} + i^{414})}^{3} = {(i^{4 \times 19 + 1} + i^{4 \times 17 + 2} + i^{4 \times 21 + 3} + i^{4 \times 103 + 2})}^{3} = [\{{(i^{4})}^{19} \times i\} + \{{(i^{4})}^{17} \times i^{2}\} + \{{(i^{4})}^{21} \times i^{3}\} + \{{(i^{4})}^{103} \times i^{2}\}] = {(i - 1 - i - 1)}^{3} (∵ i^{4} = 1, i^{3} = - i and i^{2} = - 1) = {(- 2)}^{3} = - 8 (vii) i^{30} + i^{40} + i^{60} = i^{4 \times 7 + 2} + i^{4 \times 10} + i^{4 \times 15} = [{(i^{4})}^{7} \times i^{2}] + [{(i^{4})}^{10}] + [{(i^{4})}^{15}] = - 1 + 1 + 1 (∵ i^{4} = 1, i^{2} = - 1) = 1 (viii) i^{49} + i^{68} + i^{89} + i^{110} = i^{4 \times 12 + 1} + i^{4 \times 17} + i^{4 \times 22 + 1} + i^{4 \times 27 + 2} = [{(i^{4})}^{12} \times i] + [{(i^{4})}^{17}] + [{(i^{4})}^{22} \times i] + [{(i^{4})}^{27} \times i^{2}] = i + 1 + i - 1 (∵ i^{4} = 1, i^{2} = - 1) = 2 i$

Page No 13.31:

Question 1:

Express the following complex numbers in the standard form a + i b:
(i) $(1 + i) (1 + 2 i)$
(ii) $\frac{3 + 2 i}{- 2 + i}$
(iii) $\frac{1}{(2 + i)^{2}}$
(iv) $\frac{1 - i}{1 + i}$
(v) $\frac{(2 + i)^{3}}{2 + 3 i}$
(vi) $\frac{(1 + i) (1 + \sqrt{3} i)}{1 - i}$
(vii) $\frac{2 + 3 i}{4 + 5 i}$
(viii) $\frac{(1 - i)^{3}}{1 - i^{3}}$
(ix) $(1 + 2 i)^{- 3}$
(x) $\frac{3 - 4 i}{(4 - 2 i) (1 + i)}$
(xi) $(\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{1 - 4 i}{5 + i})$
(xii) $\frac{5 + \sqrt{2} i}{1 - 2 \sqrt{i}}$

Answer:

$(i) (1 + i) (1 + 2 i) = 1 + 2 i + i + 2 i^{2} = 1 + 3 i - 2 (∵ i^{2} = - 1) = - 1 + 3 i (ii) \frac{3 + 2 i}{- 2 + i} = \frac{3 + 2 i}{- 2 + i} \times \frac{- 2 - i}{- 2 - i} = \frac{- 6 - 3 i - 4 i - 2 i^{2}}{4 - i^{2}} (∵ i^{2} = - 1) = \frac{- 6 - 7 i + 2}{4 + 1} = \frac{- 4 - 7 i}{5} = \frac{- 4}{5} - \frac{7}{5} i (iii) \frac{1}{{(2 + i)}^{2}} = \frac{1}{4 + i^{2} + 4 i} (∵ i^{2} = - 1) = \frac{1}{3 + 4 i} = \frac{1}{3 + 4 i} \times \frac{3 - 4 i}{3 - 4 i} = \frac{3 - 4 i}{9 - 16 i^{2}} = \frac{3 - 4 i}{9 + 16} = \frac{3}{25} - \frac{4}{25} i$

$(iv) \frac{1 - i}{1 + i} = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{1 + i^{2} - 2 i}{1 - i^{2}} (∵ i^{2} = - 1) = \frac{- 2 i}{2} = - i (v) \frac{{(2 + i)}^{3}}{2 + 3 i} = \frac{(4 + i^{2} + 4 i) (2 + i)}{2 + 3 i} (∵ i^{2} = - 1) = \frac{8 + 2 i^{2} + 8 i + 4 i + i^{3} + 4 i^{2}}{2 + 3 i} = \frac{2 + 11 i}{2 + 3 i} \times \frac{2 - 3 i}{2 - 3 i} = \frac{4 - 6 i + 22 i - 33 i^{2}}{4 - 9 i^{2}} = \frac{37 + 16 i}{4 + 9} = \frac{37}{13} + \frac{16}{13} i (vi) \frac{(1 + i) (1 + \sqrt{3 i})}{1 - i} = \frac{(1 + i) (1 + \sqrt{3 i})}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(1 + \sqrt{3} i) (1 + i^{2} + 2 i)}{1 - i^{2}} (∵ i^{2} = - 1) = \frac{(1 + \sqrt{3} i) 2 i}{2} = i (1 + \sqrt{3} i) = i + \sqrt{3} i^{2} = - \sqrt{3} + i (vii) \frac{2 + 3 i}{4 + 5 i} = \frac{2 + 3 i}{4 + 5 i} \times \frac{4 - 5 i}{4 - 5 i} = \frac{8 - 10 i + 12 i - 15 i^{2}}{16 - 25 i^{2}} (∵ i^{2} = - 1) = \frac{23 + 2 i}{16 + 25} = \frac{23}{41} + \frac{2}{41} i$

$(viii) \frac{{(1 - i)}^{3}}{1 - i^{3}} = \frac{(1 + i^{2} - 2 i) (1 - i)}{1 - i^{3}} (∵ i^{2} = - 1) = \frac{- 2 i (1 - i)}{1 - i^{3}} \times \frac{1 + i^{3}}{1 + i^{3}} = \frac{- 2 i (1 + i^{3} - i - i^{4})}{1 - i^{6}} = \frac{- 2 i (1 - i - i - 1)}{1 - i^{2}} = \frac{- 2 i (- 2 i)}{2} = - 2 + 0 i$

$(ix) {(1 + 2 i)}^{- 3} = \frac{1}{{(1 + 2 i)}^{3}} = \frac{1}{1 + 8 i^{3} + 6 i + 12 i^{2}} = \frac{1}{1 - 8 i + 6 i - 12} (∵ i^{2} = - 1 & i^{3} = - i) = \frac{1}{- 2 i - 11} = \frac{1}{- 2 i - 11} \times \frac{- 2 i + 11}{- 2 i + 11} = \frac{- 2 i + 11}{4 i^{2} - 121} = \frac{- 2 i + 11}{- 4 - 121} = \frac{- 2 i + 11}{- 125} = - \frac{11}{125} + \frac{2 i}{125}$

$(x) \frac{3 - 4 i}{(4 - 2 i) (1 + i)} = \frac{3 - 4 i}{4 + 2 i - 2 i^{2}} (∵ i^{2} = - 1) = \frac{3 - 4 i}{6 + 2 i} = \frac{3 - 4 i}{6 + 2 i} \times \frac{6 - 2 i}{6 - 2 i} = \frac{18 - 6 i - 24 i + 8 i^{2}}{36 - 4 i^{2}} = \frac{18 - 30 i - 8}{36 + 4} = \frac{10 - 30 i}{40} = \frac{1}{4} - \frac{3}{4} i$

$(xi) (\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{3 - 4 i}{5 + i}) = (\frac{1 + i - 2 + 8 i}{(1 - 4 i) (1 + i)}) (\frac{3 - 4 i}{5 + i}) = (\frac{- 1 + 9 i}{1 - 3 i + 4 i^{2}}) (\frac{3 - 4 i}{5 + i}) = (\frac{- 1 + 9 i}{5 - 3 i}) (\frac{3 - 4 i}{5 + i}) (∵ i^{2} = - 1) = \frac{- 3 + 4 i + 27 i - 36 i^{2}}{25 + 5 i - 15 i - 3 i^{2}} = \frac{33 + 31 i}{28 - 10 i} = \frac{33 + 31 i}{28 - 10 i} \times \frac{28 + 10 i}{28 + 10 i} = \frac{924 + 330 i + 868 i + 310 i^{2}}{784 - 100 i^{2}} = \frac{614 + 1198 i}{884} = \frac{307}{442} + \frac{599}{442} i$

$(xii) \frac{5 + \sqrt{2} i}{1 - \sqrt{2} i} = \frac{5 + \sqrt{2} i}{1 - \sqrt{2} i} \times \frac{1 + \sqrt{2} i}{1 + \sqrt{2} i} = \frac{5 + 5 \sqrt{2} i + \sqrt{2} i + 2 i^{2}}{1 - 2 i^{2}} = \frac{5 + 6 \sqrt{2} i - 2}{1 + 2} (∵ i^{2} = - 1) = \frac{3 + 6 \sqrt{2} i}{3} = 1 + 2 \sqrt{2} i$

Page No 13.31:

Question 2:

Find the real values of x and y, if
(i) $(x + i y) (2 - 3 i) = 4 + i$
(ii) $(3 x - 2 i y) (2 + i)^{2} = 10 (1 + i)$
(iii) $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i}$
(iv) $(1 + i) (x + i y) = 2 - 5 i$

Answer:

$(i) (x + i y) (2 - 3 i) = 4 + i 2 x - 3 i x + 2 i y - 3 i^{2} y = 4 + i 2 x + 3 y + i (- 3 x + 2 y) = 4 + i Comparing both the sides : 2 x + 3 y = 4 . . . . (1) - 3 x + 2 y = 1 . . . . (2) Multiplying equation (1) by 3 and equation (2) by 2 : 6 x + 9 y = 12 . . . (3) - 6 x + 4 y = 2 . . . (4) Adding equations (3) and (4) : 13 y = 14 y = \frac{14}{13} Substituting the value of y in equation (1) : 2 x + 3 \times \frac{14}{13} = 4 \Rightarrow 2 x = 4 - \frac{42}{13} \Rightarrow 2 x = \frac{10}{13} \Rightarrow x = \frac{5}{13} ∴ x = \frac{5}{13} and y = \frac{14}{13}$

$(ii) (3 x - 2 i y) {(2 + i)}^{2} = 10 (1 + i) \Rightarrow (3 x - 2 i y) (4 + i^{2} + 4 i) = 10 (1 + i) \Rightarrow (3 x - 2 i y) (3 + 4 i) = 10 (1 + i) \Rightarrow 9 x + 12 x i - 6 i y - 8 i^{2} y = 10 + 10 i \Rightarrow 9 x + 8 y + i (12 x - 6 y) = 10 + 10 i Comparing both the sides : 9 x + 8 y = 10 . . . . (1) 12 x - 6 y = 10 or, 6 x - 3 y = 5 . . . (2) Multiplying equation (1) by 3 and equation (2) by 8, 27 x + 24 y = 30 . . . . (3) 48 x - 24 y = 40 . . . . (4) Adding equations (3) and (4) : 75 x = 70 ∴ x = \frac{14}{15} Substituting the value of x in equation (1) : 9 \times \frac{14}{15} + 8 y = 10 \Rightarrow \frac{126}{15} + 8 y = 10 \Rightarrow 8 y = 10 - \frac{126}{15} \Rightarrow 8 y = \frac{24}{15} \Rightarrow y = \frac{1}{5}$

$(iii) \frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i \Rightarrow \frac{(1 + i) (3 - i) x - 2 i (3 - i) + (2 - 3 i) (3 + i) y + i (3 + i)}{(3 + i) (3 - i)} = i \Rightarrow \frac{3 x - i x + 3 i x - i^{2} x - 6 i + 2 i^{2} + 6 y + 2 i y - 9 i y - 3 i^{2} y + 3 i + i^{2}}{9 - i^{2}} = i \Rightarrow \frac{4 x + 2 i x - 3 i + 9 y - 7 i y - 3}{10} = i \Rightarrow (4 x + 9 y - 3) + i (2 x - 3 - 7 y) = 10 i Comparing both the sides : 4 x + 9 y - 3 = 0 \Rightarrow 4 x + 9 y = 3 . . . . (1) 2 x - 3 - 7 y = 10 \Rightarrow 2 x - 7 y = 13 . . . (2) Multiplying equation (2) by 2 : 4 x - 14 y = 26 . . . (3) Subtracting equation (3) from (1) : 4 x + 9 y = 3 4 x - 14 y = 26 - + - 23 y = - 23 ∴ y = - 1 Substituting the value of y in equation (1) : 4 x - 9 = 3 \Rightarrow 4 x = 12 \Rightarrow x = 3 ∴ x = 3 and y = - 1$

$(iv) (1 + i) (x + i y) = 2 - 5 i \Rightarrow x + i y + i x + i^{2} y = 2 - 5 i \Rightarrow x + i y + i x - y = 2 - 5 i \Rightarrow (x - y) + i (y + x) = 2 - 5 i Comparing both the sides, x - y = 2 . . . (1) x + y = - 5 . . . (2) Adding equations (1) and (2), 2 x = - 3 \Rightarrow x = \frac{- 3}{2} Substituting the value of x in equation (1), \frac{- 3}{2} - y = 2 \Rightarrow y = \frac{- 3}{2} - 2 \Rightarrow y = \frac{- 7}{2} ∴ x = \frac{- 3}{2} and y = \frac{- 7}{2}$

Page No 13.31:

Question 3:

Find the conjugates of the following complex numbers:
(i) 4 − 5 i
(ii) $\frac{1}{3 + 5 i}$
(iii) $\frac{1}{1 + i}$
(iv) $\frac{(3 - i)^{2}}{2 + i}$
(v) $\frac{(1 + i) (2 + i)}{3 + i}$
(vi) $\frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)}$

Answer:

$(i) Let z = 4 - 5 i ∴ \bar{z} = 4 + 5 i (z = a + i b, so \bar{z} = a - i b) (ii) Let z = \frac{1}{3 + 5 i} = \frac{1}{3 + 5 i} \times \frac{3 - 5 i}{3 - 5 i} = \frac{3 - 5 i}{9 - 25 i^{2}} = \frac{3 - 5 i}{9 + 25} = \frac{3 - 5 i}{34} ∴ \bar{z} = \frac{3 + 5 i}{34} (iii) Let z = \frac{1}{1 + i} = \frac{1}{1 + i} \times \frac{1 - i}{1 - i} = \frac{1 - i}{1 - i^{2}} = \frac{1 - i}{2} \Rightarrow \bar{z} = \frac{1 + i}{2} (iv) Let z = \frac{{(3 - i)}^{2}}{2 + i} = \frac{(9 - 6 i - 1)}{2 + i} = \frac{8 - 6 i}{2 + i} \times \frac{2 - i}{2 - i} = \frac{16 - 8 i - 12 i + 6 i^{2}}{4 - i^{2}} = \frac{10 - 20 i}{5} = 2 - 4 i ∴ z = 2 + 4 i (v) Let z = \frac{(1 + i) (2 + i)}{3 + i} = \frac{2 + i + 2 i + i^{2}}{3 + i} = \frac{1 + 3 i}{3 + i} = \frac{1 + 3 i}{3 + i} \times \frac{3 - i}{3 - i} = \frac{3 - i + 9 i - 3 i^{2}}{9 - i^{2}} = \frac{6 + 8 i}{10} = \frac{3 + 4 i}{5} \Rightarrow \bar{z} = \frac{3 - 4 i}{5} (vi) Let z = \frac{(3 - 2 i) (2 + 3 i)}{(1 + 2 i) (2 - i)} = \frac{6 + 9 i - 4 i - 6 i^{2}}{2 - i + 4 i - 2 i^{2}} = \frac{6 + 6 + 5 i}{2 + 2 + 3 i} = \frac{12 + 5 i}{4 + 3 i} \times \frac{4 - 3 i}{4 - 3 i} = \frac{48 - 36 i + 20 i - 15 i^{2}}{16 - 9 i^{2}} = \frac{63 - 16 i}{25} ∴ z = \frac{63 + 16 i}{25}$

Page No 13.32:

Question 4:

Find the multiplicative inverse of the following complex numbers:
(i) 1 − i
(ii) $(1 + i \sqrt{3})^{2}$
(iii) 4 − 3i
(iv) $\sqrt{5} + 3 i$

Answer:

$(i) Let z = 1 - i . Then, \frac{1}{z} = \frac{1}{1 - i} = \frac{1}{1 - i} \times \frac{1 + i}{1 + i} = \frac{1 + i}{1 - i^{2}} = \frac{1}{2} (1 + i) = \frac{1}{2} + \frac{1}{2} i (ii) z = {(1 + \sqrt{3} i)}^{2} = 1 + 3 i^{2} + 2 \sqrt{3} i = - 2 + 2 \sqrt{3} i Then, \frac{1}{z} = \frac{1}{- 2 + 2 \sqrt{3} i} \times \frac{- 2 - 2 \sqrt{3} i}{- 2 - 2 \sqrt{3} i} = \frac{- 2 - 2 \sqrt{3} i}{4 - 12 i^{2}} = \frac{- 2 - 2 \sqrt{3} i}{16} = \frac{- 1}{8} - \frac{\sqrt{3}}{8} i (iii) z = 4 - 3 i Then, \frac{1}{z} = \frac{1}{4 - 3 i} \times \frac{4 + 3 i}{4 + 3 i} = \frac{4 + 3 i}{16 - 9 i^{2}} = \frac{4 + 3 i}{25} = \frac{4}{25} + \frac{3}{25} i (iv) z = \sqrt{5} + 3 i Then, \frac{1}{z} = \frac{1}{\sqrt{5} + 3 i} \times \frac{\sqrt{5} - 3 i}{\sqrt{5} - 3 i} = \frac{\sqrt{5} - 3 i}{5 - 9 i^{2}} = \frac{\sqrt{5} - 3 i}{5 + 9} = \frac{\sqrt{5} - 3 i}{14} = \frac{\sqrt{5}}{14} - \frac{3}{14} i$

Page No 13.32:

Question 5:

If $z_{1} = 2 - i, z_{2} = 1 + i, find |\frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i}|$

Answer:

$Given : z_{1} = 2 - i, z_{2} = 1 + i ∴ |\frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i}| = |\frac{2 - i + 1 + i + 1}{2 - i - 1 - i + i}| = |\frac{4}{1 - i}| = \frac{4}{|1 - i|} Also, |1 - i| = \sqrt{1^{2} + i^{2}} (∵ |a + b i| = \sqrt{a^{2} + b^{2}}) = \sqrt{2} ∴ |\frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i}| = \frac{4}{\sqrt{2}}$

Page No 13.32:

Question 6:

If $z_{1} = 2 - i, z_{2} = - 2 + i,$ find
(i) Re $(\frac{z_{1} z_{2}}{z_{1}})$
(ii) Im $(\frac{1}{z_{1} z_{1}})$

Answer:

$(i) z_{1} = 2 - i, z_{2} = - 2 + i, z_{1} = 2 + i ∴ (\frac{z_{1} z_{2}}{z_{1}}) = (\frac{[2 - i] [- 2 + i]}{2 + i}) = (\frac{- 4 + 2 i + 2 i - i^{2}}{2 + i}) = (\frac{- 3 + 4 i}{2 + i}) = [\frac{- 3 + 4 i}{2 + i} \times (\frac{2 - i}{2 - i})] = (\frac{- 6 + 3 i + 8 i - 4 i^{2}}{2^{2} - i^{2}}) = (\frac{- 2 + 11 i}{4 - (- 1)}) = (\frac{- 2 + 11 i}{5}) Re (\frac{z_{1} z_{2}}{z_{1}}) = \frac{- 2}{5} (ii) (\frac{1}{z_{1} z_{1}}) = \frac{1}{(2 - i) (2 + i)} = \frac{1}{2^{2} - i^{2}} = \frac{1}{5} Im (\frac{1}{z_{1} z_{1}}) = 0 (Since no term containing i is present)$

Page No 13.32:

Question 7:

Find the modulus of $\frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}$

Answer:

$\frac{1 + i}{1 - i} - \frac{1 - i}{1 + i} = \frac{(1 + i) (1 + i) - (1 - i) (1 - i)}{(1 - i) (1 + i)} = \frac{1 + i^{2} + 2 i - 1 - i^{2} + 2 i}{1^{2} - i^{2}} = \frac{4 i}{2} (∵ i^{2} = - 1) = 2 i ∴ |2 i| = \sqrt{0^{2} + 2^{2}} = 2 (∵ |a + b i| = \sqrt{a^{2} + b^{2}}) \Rightarrow |\frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}| = 2$

Page No 13.32:

Question 8:

If $x + i y = \frac{a + i b}{a - i b}$ , prove that x² + y² = 1

Answer:

$x + i y = \frac{a + i b}{a - i b} Taking \mod on both the sides : |x + i y| = |\frac{a + i b}{a - i b}| \Rightarrow \sqrt{x^{2} + y^{2}} = \frac{\sqrt{a^{2} + b^{2}}}{\sqrt{a^{2} + b^{2}}} \Rightarrow \sqrt{x^{2} + y^{2}} = 1 \Rightarrow x^{2} + y^{2} = 1 Hence proved .$

Page No 13.32:

Question 9:

Find the least positive integral value of n for which ${(\frac{1 + i}{1 - i})}^{n}$ is real.

Answer:

${(\frac{1 + i}{1 - i})}^{n} = {[\frac{1 + i}{1 - i} \times (\frac{1 + i}{1 + i})]}^{n} = {(\frac{1 + i^{2} + 2 i}{1 - i^{2}})}^{n} = {(\frac{1 - 1 + 2 i}{1 + 1})}^{n} = {(\frac{2 i}{2})}^{n} = i^{n} For i^{n} to be real, the least positive value of n will be 2 . As i^{2} = - 1$

Page No 13.32:

Question 10:

Find the real values of θ for which the complex number $\frac{1 + i cosθ}{1 - 2 i cosθ}$ is purely real.

Answer:

$\frac{1 + i \cos θ}{1 - 2 i \cos θ} = \frac{1 + i \cos θ}{1 - 2 i \cos θ} \times \frac{1 + 2 i \cos θ}{1 + 2 i \cos θ} = \frac{1 + 2 i \cos θ + i \cos θ - 2 \cos θ}{1 + 4 \cos^{2} θ} = \frac{1 - 2 \cos θ + i 3 \cos θ}{1 + 4 \cos^{2} θ} For it to be purely real, the imaginary part must be zero . 3 \cos θ = 0 This is true for odd multiples of \frac{π}{2} . ∴ θ = (2 n + 1) \frac{π}{2}, n \in Z$

Page No 13.32:

Question 11:

Find the smallest positive integer value of m for which $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is a real number.

Answer:

$\frac{{(1 + i)}^{m}}{{(1 - i)}^{m - 2}} = \frac{{(1 + i)}^{m}}{{(1 - i)}^{m}} \times {(1 - i)}^{2} = {(\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i})}^{m} \times (1 + i^{2} - 2 i) = {(\frac{1 + i^{2} + 2 i}{1 - i^{2}})}^{m} \times (1 - 1 - 2 i) = {(\frac{1 - 1 + 2 i}{1 + 1})}^{m} \times (- 2 i) = - 2 i (i^{m}) = - 2 {(i)}^{m + 1} For this to be real, the smallest positive value of m will be 1 . Thus, i^{1 + 1} = i^{2} = - 1, which is real .$

Page No 13.32:

Question 12:

If ${(\frac{1 + i}{1 - i})}^{3} - {(\frac{1 - i}{1 + i})}^{3} = x + i y$ , find (x, y).

Answer:

$(\frac{1 + i}{1 - i}) = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{{(1 + i)}^{2}}{1^{2} - i^{2}} = \frac{1^{2} + i^{2} + 2 i}{1 + 1} [∵ i^{2} = - 1] = \frac{1 - 1 + 2 i}{2} = \frac{2 i}{2} = i . . . . (1)$

Also,
$(\frac{1 - i}{1 + i}) = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{{(1 - i)}^{2}}{1^{2} - i^{2}} = \frac{1^{2} + i^{2} - 2 i}{1 + 1} [∵ i^{2} = - 1] = \frac{1 - 1 - 2 i}{2} = \frac{- 2 i}{2} = - i . . . . (2)$

It is given that,
${(\frac{1 + i}{1 - i})}^{3} - {(\frac{1 - i}{1 + i})}^{3} = x + i y \Rightarrow (i)^{3} - (- i)^{3} = x + i y [From (1) and (2)] \Rightarrow i^{3} + i^{3} = x + i y \Rightarrow 2 i^{3} = x + i y \Rightarrow 0 - 2 i = x + i y [∵ i^{3} = - i] \Rightarrow x = 0 and y = - 2$

Thus, (x, y) = (0, −2).

Page No 13.32:

Question 13:

If $\frac{{(1 + i)}^{2}}{2 - i} = x + i y$ , find x + y.

Answer:

$\frac{{(1 + i)}^{2}}{2 - i} = \frac{1^{2} + i^{2} + 2 i}{2 - i} = \frac{1 - 1 + 2 i}{2 - i} [∵ i^{2} = - 1] = \frac{2 i}{2 - 1} \times \frac{2 + i}{2 + i} = \frac{2 i (2 + i)}{2^{2} - i^{2}} = \frac{4 i + 2 i^{2}}{4 + 1} [∵ i^{2} = - 1] = \frac{4 i - 2}{5} = \frac{- 2}{5} + \frac{4}{5} i . . . . (1)$

It is given that,
$\frac{{(1 + i)}^{2}}{2 - i} = x + i y \Rightarrow - \frac{2}{5} + \frac{4}{5} i = x + i y [From (1)] \Rightarrow x = - \frac{2}{5} and y = \frac{4}{5}$

$∴ x + y = \frac{- 2}{5} + \frac{4}{5} = \frac{2}{5}$

Thus, x + y = $\frac{2}{5}$ .

Page No 13.32:

Question 14:

If ${(\frac{1 - i}{1 + i})}^{100} = a + i b$ , find (a, b).

Answer:

$\frac{1 - i}{1 + i} = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{{(1 - i)}^{2}}{1^{2} - i^{2}} = \frac{1^{2} + i^{2} - 2 i}{1 + 1} [∵ i^{2} = - 1] = \frac{1 - 1 - 2 i}{2} = \frac{- 2 i}{2} = - i . . . . (1)$

It is given that,
${(\frac{1 - i}{1 + i})}^{100} = a + i b \Rightarrow (- i)^{100} = a + i b [From (1)] \Rightarrow i^{4 \times 25} = a + i b \Rightarrow 1 + 0 i = a + i b [∵ i^{4} = 1] \Rightarrow a = 1 and b = 0$

Thus, (a, b) = (1, 0).

Page No 13.32:

Question 15:

If $a = \cos θ + i \sin θ$ , find the value of $\frac{1 + a}{1 - a}$ .

Answer:

$\frac{1 + a}{1 - a} = \frac{1 + \cos θ + i \sin θ}{1 - \cos θ - i \sin θ} = \frac{(1 + c o s θ) + i s i n θ}{(1 - c o s θ) - i s i n θ} \times \frac{(1 - c o s θ) + i s i n θ}{(1 - c o s θ) + i s i n θ} = \frac{1 - \cos θ + i \sin θ + \cos θ - \cos^{2} θ + i \cos θ \sin θ + i \sin θ - i \sin θ \cos θ + i^{2} \sin^{2} θ}{(1 - \cos θ)^{2} - i^{2} \sin^{2} θ} = \frac{1 - \cos^{2} θ - \sin^{2} θ + 2 i \sin θ}{1 + \cos^{2} θ - 2 i \cos θ + \sin^{2} θ} [∵ i^{2} = - 1] = \frac{\sin^{2} θ - \sin^{2} θ + 2 i \sin θ}{2 - 2 i \cos θ} [∵ \cos^{2} θ + \sin^{2} θ = 1] = \frac{i \sin θ}{1 - \cos θ} = \frac{2 i \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} = 2 i \cot \frac{θ}{2}$

Thus, $\frac{1 + a}{1 - a} = 2 i \cot \frac{θ}{2}$ .

Page No 13.32:

Question 16:

Evaluate the following:
(i) $2 x^{3} + 2 x^{2} - 7 x + 72, when x = \frac{3 - 5 i}{2}$
(ii) $x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44, when x = 3 + 2 i$
(iii) $x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9, when x = - 1 + i \sqrt{2}$
(iv) $x^{6} + x^{4} + x^{2} + 1, when x = \frac{1 + i}{\sqrt{2}}$
(v) $2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41, when x = - 2 - \sqrt{3} i$

Answer:

$(i) x = \frac{3 - 5 i}{2} \Rightarrow x^{2} = {(\frac{3 - 5 i}{2})}^{2} = \frac{9 + 25 i^{2} - 30 i}{4} = \frac{- 16 - 30 i}{4} \Rightarrow x^{3} = \frac{- 16 - 30 i}{4} \times \frac{3 - 5 i}{2} = \frac{- 48 + 80 i - 90 i + 150 i^{2}}{8} = \frac{- 198 - 10 i}{8} ∴ 2 x^{3} + 2 x^{2} - 7 x + 72 = 2 (\frac{- 198 - 10 i}{8}) + 2 (\frac{- 16 - 30 i}{4}) - 7 (\frac{3 - 5 i}{2}) + 72 = \frac{- 198 - 10 i - 32 - 60 i - 42 + 70 i + 288}{4} = \frac{16}{4} = 4 (ii) x = 3 + 2 i \Rightarrow x^{2} = {(3 + 2 i)}^{2} = 9 + 4 i^{2} + 12 i = 5 + 12 i \Rightarrow x^{3} = x^{2} \times x = (5 + 12 i) \times (3 + 2 i) = 15 + 10 i + 36 i - 24 = - 9 + 46 i \Rightarrow x^{4} = {(x^{2})}^{2} = {(5 + 12 i)}^{2} = 25 + 144 i^{2} + 120 i = - 119 + 120 i \Rightarrow x^{4} - 4 x^{3} + 4 x^{2} + 8 x + 44 = - 119 + 120 i - 4 (- 9 + 46 i) + 4 (5 + 12 i) + 8 (3 + 2 i) + 44 = - 119 + 120 i + 36 - 184 i + 20 + 48 i + 24 + 16 i + 44 = 5 (iii) x = - 1 + \sqrt{2} i \Rightarrow x^{2} = {(- 1 + \sqrt{2} i)}^{2} = 1 + 2 i^{2} - 2 \sqrt{2} i = - 1 - 2 \sqrt{2} i \Rightarrow x^{3} = (- 1 - 2 \sqrt{2} i) \times (- 1 + \sqrt{2} i) = 1 - \sqrt{2} i + 2 \sqrt{2} i - 4 i^{2} = 5 + \sqrt{2} i \Rightarrow x^{4} = {(- 1 - 2 \sqrt{2} i)}^{2} = 1 + 8 i^{2} + 4 \sqrt{2} i = - 7 + 4 \sqrt{2} i \Rightarrow x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 9 = - 7 + 4 \sqrt{2} i + 4 (5 + \sqrt{2} i) + 6 (- 1 - 2 \sqrt{2} i) + 4 (- 1 + \sqrt{2} i) + 9 = - 7 + 4 \sqrt{2} i + 20 + 4 \sqrt{2} i - 6 - 12 \sqrt{2} i - 4 + 4 \sqrt{2} i + 9 = 12 (iv) x = \frac{1 + i}{\sqrt{2}} \Rightarrow x^{2} = {(\frac{1 + i}{\sqrt{2}})}^{2} = (\frac{1 + i^{2} + 2 i}{2}) = \frac{2 i}{2} = i \Rightarrow x^{6} = {(x^{2})}^{3} = i^{3} = - i \Rightarrow x^{2} = i \Rightarrow x^{4} = {(x^{2})}^{2} = i^{2} = - 1 Now, x^{6} + x^{4} + x^{2} + 1 = - i - 1 + i + 1 = 0$

$(v) x = - 2 - \sqrt{3} i \Rightarrow x^{2} = {(- 2 - \sqrt{3} i)}^{2} = (- 2)^{2} + (- \sqrt{3} i)^{2} + 2 (- 2) (- \sqrt{3} i) = 4 + 3 i^{2} + 4 \sqrt{3} i = 4 - 3 + 4 \sqrt{3} i [∵ i^{2} = - 1] = 1 + 4 \sqrt{3} i \Rightarrow x^{3} = (1 + 4 \sqrt{3} i) \times (- 2 - \sqrt{3} i) = - 2 - \sqrt{3} i - 8 \sqrt{3} i - 12 i^{2} = 10 - 9 \sqrt{3} i [∵ i^{2} = - 1] \Rightarrow x^{4} = {(1 + 4 \sqrt{3} i)}^{2} = 1 + 48 i^{2} + 8 \sqrt{3} i = - 47 + 8 \sqrt{3} i [∵ i^{2} = - 1] \Rightarrow 2 x^{4} + 5 x^{3} + 7 x^{2} - x + 41 = 2 (- 47 + 8 \sqrt{3} i) + 5 (10 - 9 \sqrt{3} i) + 7 (1 + 4 \sqrt{3} i) - (- 2 - \sqrt{3} i) + 41 = - 94 + 16 \sqrt{3} i + 50 - 45 \sqrt{3} i + 7 + 28 \sqrt{3} i + 2 + \sqrt{3} i + 41 = 6$

Page No 13.32:

Question 17:

For a positive integer n, find the value of $(1 - i)^{n} {(1 - \frac{1}{i})}^{n}$ .

Answer:

$(1 - i)^{n} {(1 - \frac{1}{i})}^{n} = {(1 - i)}^{n} {(1 - \frac{i^{4}}{i})}^{n} [∵ i^{4} = 1] = {(1 - i)}^{n} {(1 - i^{3})}^{n} = {(1 - i)}^{n} {(1 + i)}^{n} [∵ i^{3} = - i] = {[(1 - i) (1 + i)]}^{n} = (1 - i^{2})^{n} = 2^{n} [∵ i^{2} = - 1]$

Thus, the value of $(1 - i)^{n} {(1 - \frac{1}{i})}^{n}$ is 2ⁿ.

Page No 13.33:

Question 18:

If $(1 + i) z = (1 - i) \bar{z}$ , then show that $z = - i \bar{z}$ .

Answer:

$(1 + i) z = (1 - i) \bar{z} \Rightarrow \frac{z}{\bar{z}} = \frac{1 - i}{1 + i} \Rightarrow \frac{z}{\bar{z}} = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} \Rightarrow \frac{z}{\bar{z}} = \frac{1 + i^{2} - 2 i}{1 - i^{2}} \Rightarrow \frac{z}{\bar{z}} = \frac{1 - 1 - 2 i}{1 + 1} [∵ i^{2} = - 1] \Rightarrow \frac{z}{\bar{z}} = \frac{- 2 i}{2} \Rightarrow \frac{z}{\bar{z}} = - i \Rightarrow z = - i \bar{z}$

Hence, $z = - i \bar{z}$ .

Page No 13.33:

Question 19:

Solve the system of equations $Re (z^{2}) = 0, |z| = 2$ .

Answer:

Let $z = x + i y$ .
Then ,
$z^{2} = {(x + i y)}^{2} = x^{2} + i^{2} y^{2} + 2 i x y = x^{2} - y^{2} + 2 i x y [∵ i^{2} = - 1]$

and $|z| = \sqrt{x^{2} + y^{2}}$

According to the question,
$Re (z^{2}) = 0 and |z| = 2 \Rightarrow x^{2} - y^{2} = 0 and \sqrt{x^{2} + y^{2}} = 2 \Rightarrow x^{2} - y^{2} = 0 and x^{2} + y^{2} = 4 On Adding both the equations, we get 2 x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm \sqrt{2} \Rightarrow y^{2} = 2 \Rightarrow y = \pm \sqrt{2}$

Thus, $x = \pm \sqrt{2} and y = \pm \sqrt{2}$ .

Page No 13.33:

Question 20:

If $\frac{z - 1}{z + 1}$ is purely imaginary number ( $z \neq - 1$ ), find the value of $|z|$ .

Answer:

Let $z = x + i y$ .
Then,
$\frac{z - 1}{z + 1} = \frac{x + i y - 1}{x + i y + 1} = \frac{(x - 1) + i y}{(x + 1) + i y} \times \frac{(x + 1) - i y}{(x + 1) - i y} = \frac{x^{2} + x - i x y - x - 1 + i y + i x y + i y - i^{2} y^{2}}{{(x + 1)}^{2} - i^{2} y^{2}} = \frac{x^{2} + y^{2} - 1 + 2 i y}{x^{2} + 1 + 2 x + y^{2}} [∵ i^{2} = - 1]$

If $\frac{z - 1}{z + 1}$ is purely imaginary number, then
$Re (\frac{z - 1}{z + 1}) = 0 \Rightarrow x^{2} + y^{2} - 1 = 0 \Rightarrow x^{2} + y^{2} = 1 \Rightarrow {|z|}^{2} = 1 \Rightarrow |z| = 1$

Thus, the value of $|z|$ is 1.

Page No 13.33:

Question 21:

If z₁ is a complex number other than −1 such that $|z_{1}| = 1$ and $z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ , then show that the real parts of z₂ is zero.

Answer:

Let $z = x + i y$ .
Then,
$z_{2} = \frac{z_{1} - 1}{z_{1} + 1} = \frac{x + i y - 1}{x + i y + 1} = \frac{(x - 1) + i y}{(x + 1) + i y} \times \frac{(x + 1) - i y}{(x + 1) - i y} = \frac{x^{2} + x - i x y - x - 1 + i y + i x y + i y - i^{2} y^{2}}{{(x + 1)}^{2} - i^{2} y^{2}} = \frac{x^{2} + y^{2} - 1 + 2 i y}{x^{2} + 1 + 2 x + y^{2}} [∵ i^{2} = - 1]$

Now,
$Re (z_{2}) = \frac{x^{2} + y^{2} - 1}{x^{2} + y^{2} + 1 + 2 x} = 0 [∵ |z_{1}| = 1 \Rightarrow x^{2} + y^{2} = 1]$

Thus, the real parts of z₂ is zero.

Page No 13.33:

Question 22:

If $|z + 1| = z + 2 (1 + i)$ , find z.

Answer:

Let $z = x + i y$ .
Then,
$z + 1 = (x + 1) + i y \Rightarrow |z + 1| = \sqrt{{(x + 1)}^{2} + y^{2}}$

$∴ |z + 1| = z + 2 (1 + i) \Rightarrow \sqrt{x^{2} + 2 x + 1 + y^{2}} = (x + i y) + 2 + 2 i \Rightarrow \sqrt{x^{2} + 2 x + 1 + y^{2}} = (x + 2) + i (y + 2) \Rightarrow \sqrt{x^{2} + 2 x + 1 + y^{2}} = (x + 2) and y + 2 = 0 \Rightarrow x^{2} + 2 x + 1 + y^{2} = {(x + 2)}^{2} and y = - 2 \Rightarrow x^{2} + 2 x + 1 + y^{2} = x^{2} + 4 x + 4 and y = - 2 \Rightarrow y^{2} = 2 x + 3 and y = - 2 \Rightarrow 4 = 2 x + 3 and y = - 2 \Rightarrow 2 x = 1 and y = - 2 \Rightarrow x = \frac{1}{2} and y = - 2$

$∴ z = x + i y = \frac{1}{2} - 2 i$

Thus, $z = \frac{1}{2} - 2 i$

Page No 13.33:

Question 23:

Solve the equation $|z| = z + 1 + 2 i$ .

Answer:

Let $z = x + i y$ .
Then,
$|z| = \sqrt{x^{2} + y^{2}}$

$∴ |z| = z + 1 + 2 i \Rightarrow \sqrt{x^{2} + y^{2}} = (x + i y) + 1 + 2 i \Rightarrow \sqrt{x^{2} + y^{2}} = (x + 1) + i (y + 2) \Rightarrow \sqrt{x^{2} + y^{2}} = (x + 1) and y + 2 = 0 \Rightarrow x^{2} + y^{2} = {(x + 1)}^{2} and y = - 2 \Rightarrow x^{2} + y^{2} = x^{2} + 1 + 2 x and y = - 2 \Rightarrow y^{2} = 2 x + 1 and y = - 2 \Rightarrow 4 = 2 x + 1 and y = - 2 \Rightarrow 2 x = 3 and y = - 2 \Rightarrow x = \frac{3}{2} and y = - 2$

$∴ z = x + i y = \frac{3}{2} - 2 i$

â€‹Thus, $z = \frac{3}{2} - 2 i$

Page No 13.33:

Question 24:

What is the smallest positive integer n for which ${(1 + i)}^{2 n} = {(1 - i)}^{2 n}$ ?

Answer:

${(1 + i)}^{2 n} = {(1 - i)}^{2 n} \Rightarrow {[{(1 + i)}^{2}]}^{n} = {[{(1 - i)}^{2}]}^{n} \Rightarrow {(1^{2} + i^{2} + 2 i)}^{n} = {(1^{2} + i^{2} - 2 i)}^{n} \Rightarrow {(1 - 1 + 2 i)}^{n} = {(1 - 1 - 2 i)}^{n} [∵ i^{2} = - 1] \Rightarrow {(2 i)}^{n} = {(- 2 i)}^{n} \Rightarrow {(2 i)}^{n} = {(- 1)^{n} (2 i)}^{n} \Rightarrow (- 1)^{n} = 1 \Rightarrow n is a multiple of 2$

Thus, the smallest positive integer n for which ${(1 + i)}^{2 n} = {(1 - i)}^{2 n}$ is 2.

Page No 13.33:

Question 25:

If z₁, z₂, z₃ are complex numbers such that $|z_{1}| = |z_{2}| = |z_{3}| = |\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}}| = 1$ , then find the value of $|z_{1} + z_{2} + z_{3}|$ .

Answer:

$|z_{1} + z_{2} + z_{3}| = |\frac{z_{1} \bar{z_{1}}}{\bar{z_{1}}} + \frac{z_{2} \bar{z_{2}}}{\bar{z_{2}}} + \frac{z_{3} \bar{z_{3}}}{\bar{z_{3}}}| = |\frac{{|z_{1}|}^{2}}{\bar{z_{1}}} + \frac{{|z_{2}|}^{2}}{\bar{z_{2}}} + \frac{{|z_{3}|}^{2}}{\bar{z_{3}}}| = |\frac{1}{\bar{z_{1}}} + \frac{1}{\bar{z_{2}}} + \frac{1}{\bar{z_{3}}}| [∵ |z_{1}| = |z_{2}| = |z_{3}| = 1] = \bar{|\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}}|} = 1 [∵ |\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}}| = 1]$

Thus, the value of $|z_{1} + z_{2} + z_{3}|$ is 1.

Page No 13.33:

Question 26:

Find the number of solutions of $z^{2} + {|z|}^{2} = 0$

Answer:

Let $z = x + i y$ .
Then,
$|z| = \sqrt{x^{2} + y^{2}}$

$∴ z^{2} + {|z|}^{2} = 0 \Rightarrow {(x + i y)}^{2} + {(\sqrt{x^{2} + y^{2}})}^{2} = 0 \Rightarrow x^{2} + i^{2} y^{2} + 2 i x y + x^{2} + y^{2} = 0 \Rightarrow x^{2} - y^{2} + 2 i x y + x^{2} + y^{2} = 0 \Rightarrow 2 x^{2} + 2 i x y = 0 \Rightarrow 2 x (x + i y) = 0 \Rightarrow x = 0 or x + i y = 0 \Rightarrow x = 0 or z = 0$

For $x = 0, z = 0 + i y$

â€‹Thus, there are infinitely many solutions of the form $z = 0 + i y, y \in R$ .

Page No 13.39:

Question 1:

Find the square root of the following complex numbers:
(i) −5 + 12i
(ii) −7 − 24i
(iii) 1 − i
(iv) −8 − 6i
(v) 8 −15i
(vi) $- 11 - 60 \sqrt{- 1}$
(vii) $1 + 4 \sqrt{- 3}$
(viii) 4i
(ix) −i

Answer:

$\sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} + i \sqrt{\frac{|z| - Re (z)}{2}}], if Im (z) > 0 \sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} - i \sqrt{\frac{|z| - Re (z)}{2}}], if Im (z) < 0 (i) z = - 5 + 12 i, Re (z) = - 5 and |z| = \sqrt{25 + 144} = 13 Here, Im (z) > 0 ∴ \sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} + i \sqrt{\frac{|z| - Re (z)}{2}}] = \pm [\sqrt{\frac{13 - 5}{2}} + i \sqrt{\frac{13 + 5}{2}}] = \pm [\sqrt{4} + i \sqrt{9}] = \pm (2 + 3 i) (ii) z = - 7 - 24 i, Re (z) = - 7, |z| = \sqrt{49 + 576} = 25 Here, Im (z) < 0 \sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} - i \sqrt{\frac{|z| - Re (z)}{2}}] = \pm [\sqrt{\frac{25 - 7}{2}} - i \sqrt{\frac{25 + 7}{2}}] = \pm (3 - 4 i) (iii) z = 1 - i, Re (z) = 1, |z| = \sqrt{1 + 1} = \sqrt{2} Here, Im (z) < 0 ∴ \sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} - i \sqrt{\frac{|z| - Re (z)}{2}}] = \pm [\sqrt{\frac{\sqrt{2} + 1}{2}} - i \sqrt{\frac{\sqrt{2} - 1}{2}}] (iv) z = - 8 - 6 i, Re (z) = - 8, |z| = \sqrt{64 + 36} = 10 Here, Im (z) < 0 ∴ \sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} - i \sqrt{\frac{|z| - Re (z)}{2}}] = \pm [\sqrt{\frac{10 - 8}{2}} - i \sqrt{\frac{10 + 8}{2}}] = \pm (1 - 3 i) (v) z = 8 - 15 i, Re (z) = 8, |z| = \sqrt{64 + 225} = 17 Here, Im (z) < 0 ∴ \sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} - i \sqrt{\frac{|z| - Re (z)}{2}}] = \pm [\sqrt{\frac{17 + 8}{2}} - i \sqrt{\frac{17 + 8}{2}}] = \pm \frac{1}{\sqrt{2}} (5 - 3 i) (vi) - 11 - 60 \sqrt{- 1} = - 11 - 60 i, Re (z) = - 11, |z| = \sqrt{121 + 3600} = 61 Here, Im (z) < 0 ∴ \sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} - i \sqrt{\frac{|z| - Re (z)}{2}}] = \pm [\sqrt{\frac{61 - 11}{2}} - i \sqrt{\frac{61 + 11}{2}}] = \pm (5 - 6 i) (vii) z = 1 + 4 \sqrt{3} \sqrt{- 1} = 1 + 4 \sqrt{3} i, Re (z) = 1, |z| = \sqrt{1 + 16 \times 3} = 7 Here, Im (z) > 0 ∴ \sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} + i \sqrt{\frac{|z| - Re (z)}{2}}] = \pm [\sqrt{\frac{7 + 1}{2}} + i \sqrt{\frac{7 - 1}{2}}] = \pm (2 + \sqrt{3} i) (viii) z = 0 + 4 i, Re (z) = 0, |z| = 4 Here, Im (z) > 0 ∴ \sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} + i \sqrt{\frac{|z| - Re (z)}{2}}] = \pm [\sqrt{\frac{4 + 0}{2}} + i \sqrt{\frac{4 - 0}{2}}] = \pm (\sqrt{2} + i \sqrt{2}) = \pm \sqrt{2} (1 + i) (ix) z = - i, Re (z) = 0, |z| = 1 Here, Im (z) < 0 ∴ \sqrt{z} = \pm [\sqrt{\frac{|z| + Re (z)}{2}} - i \sqrt{\frac{|z| - Re (z)}{2}}] = \pm [\sqrt{\frac{1}{2}} - i \sqrt{\frac{1}{2}}] = \pm \frac{1}{\sqrt{2}} (1 - i)$

Page No 13.4:

Question 2:

Show that 1 + i¹⁰ + i²⁰ + i³⁰ is a real number.

Answer:

$1 + i^{10} + i^{20} + i^{30} = 1 + i^{4 \times 2 + 2} + i^{4 \times 5} + i^{4 \times 7 + 2} = 1 + [{(i^{4})}^{2} \times i^{2}] + {(i^{4})}^{5} + [{(i^{4})}^{7} \times i^{2}] = 1 + i^{2} + 1 + i^{2} (∵ i^{4} = 1) = 1 - 1 + 1 - 1 (∵ i^{2} = - 1) = 0 This is a real number . Hence proved .$

Page No 13.4:

Question 3:

Find the values of the following expressions:
(i) i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰
(ii) i³⁰ + i⁸⁰ + i¹²⁰
(iii) i + i² + i³ + i⁴
(iv) i⁵ + i¹⁰ + i¹⁵
(v) $\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}}$
(vi) 1+ i² + i⁴ + i⁶ + i⁸ + ... + i²⁰
(vii) (1 + i)⁶ + (1 − i)³

Answer:

$(i) i^{49} + i^{68} + i^{89} + i^{110} = i^{4 \times 12 + 1} + i^{4 \times 17} + i^{4 \times 22 + 1} + i^{4 \times 27 + 2} = \{{(i^{4})}^{12} \times i\} + \{{(i^{4})}^{17}\} + \{{(i^{4})}^{22} \times i\} + \{{(i^{4})}^{27} \times i^{2}\} = i + 1 + i + i^{2} [∵ i^{4} = 1] = 2 i + 1 - 1 [∵ i^{2} = - 1] = 2 i (ii) i^{30} + i^{80} + i^{120} = i^{4 \times 7 + 2} + i^{4 \times 20} + i^{4 \times 30} = \{{(i^{4})}^{7} \times i^{2}\} + \{{(i^{4})}^{20}\} + \{{(i^{4})}^{30}\} = i^{2} + 1 + 1 [∵ i^{4} = 1] = - 1 + 2 [∵ i^{2} = - 1] = 1 (iii) i + i^{2} + i^{3} + i^{4} = i - 1 - i + 1 [∵ i^{2} = - 1, i^{3} = - i and i^{4} = 1] = 0 (iv) i^{5} + i^{10} + i^{15} = i^{4 \times 1 + 1} + i^{4 \times 2 + 2} + i^{4 \times 3 + 3} = \{{(i^{4})}^{1} \times i\} + \{{(i^{4})}^{2} \times i^{2}\} + \{{(i^{4})}^{3} \times i^{3}\} = i + i^{2} + i^{3} [∵ i^{4} = 1] = i - 1 - i [∵ i^{2} = - 1, i^{3} = - i] = - 1 (v) \frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} = \frac{i^{4 \times 148} + i^{4 \times 147 + 2} + i^{4 \times 147} + i^{4 \times 146 + 2} + i^{4 \times 146}}{i^{4 \times 145 + 2} + i^{4 \times 145} + i^{4 \times 144 + 2} + i^{4 \times 144} + i^{4 \times 143 + 2}} = \frac{{(i^{4})}^{148} + \{{(i^{4})}^{147} \times i^{2}\} + \{{(i^{4})}^{146}\} + \{{(i^{4})}^{146} \times i^{2}\} + \{{(i^{4})}^{146}\}}{\{{(i^{4})}^{145} \times i^{2}\} + \{{(i^{4})}^{145}\} + \{{(i^{4})}^{144} \times i^{2}\} + \{{(i^{4})}^{144}\} + \{{(i^{4})}^{143} \times i^{2}\}} = \frac{1 + i^{2} + 1 + i^{2} + 1}{i^{2} + 1 + i^{2} + 1 + i^{2}} [∵ i^{4} = 1] = \frac{1 - 1 + 1 - 1 + 1}{- 1 + 1 - 1 + 1 - 1} [∵ i^{2} = - 1] = - 1$

$(vi) 1 + i^{2} + i^{4} + i^{6} + i^{8} + . . . + i^{20} ∵ i^{2} = - 1, i^{4} = 1, i^{6} = - 1, i^{8} = 1, . . . . . i^{20} = 1 ∴ 1 + i^{2} + i^{4} + i^{6} + i^{8} + . . . + i^{20} = [1 + (- 1)] + [1 + (- 1)] + [1 + (- 1)] + . . . + [1 + (- 1)] + 1 = 5 \times [1 + (- 1)] + 1 [As, there are 11 terms] = 5 \times 0 + 1 = 1$

(vii) (1 + i)⁶ + (1 − i)³
= [(1 + i)²]³ + (1 − i)³
= [1² + i² + 2i]³ + (1³ − i³ + 3i²− 3i)
= [1 − 1 + 2i]³ + (1 + i − 3− 3i) [âˆµ i² = −1, i³= −i]
= (2i)³ + (−2− 2i)
= 8i³ − 2− 2i
= −8i − 2− 2i [âˆµ i³= −i]
= −10i − 2

Page No 13.57:

Question 1:

Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
(i) 1 + i
(ii) $\sqrt{3} + i$
(iii) 1 − i
(iv) $\frac{1 - i}{1 + i}$
(v) $\frac{1}{1 + i}$
(vi) $\frac{1 + 2 i}{1 - 3 i}$
(vii) $\sin 120 ° - i \cos 120 °$
(viii) $\frac{- 16}{1 + i \sqrt{3}}$

Answer:

$(i) z = 1 + i r = |z| = \sqrt{1 + 1} = \sqrt{2} Let \tan α = |\frac{Im (z)}{Re (z)}| \Rightarrow \tan α = (\frac{1}{1}) \Rightarrow α = \frac{π}{4} Since point (1, 1) lies in the first quadrant, the argument of z is given by θ = α = \frac{π}{4} Polar form = r (\cos θ + i \sin θ) = \sqrt{2} (\cos \frac{π}{4} + i \sin \frac{π}{4}) (ii) z = \sqrt{3} + i r = |z| = \sqrt{3 + 1} = \sqrt{4} = 2 Let \tan α = |\frac{Im (z)}{Re (z)}| \Rightarrow \tan α = (\frac{1}{\sqrt{3}}) \Rightarrow α = \frac{π}{6} Since point (\sqrt{3}, 1) lies in the first quadrant, the argument of z is given by θ = α = \frac{π}{6}$

$Polar form = r (\cos θ + i \sin θ) = 2 (\cos \frac{π}{6} + i \sin \frac{π}{6})$

$(iii) z = 1 - i r = |z| = \sqrt{1 + 1} = \sqrt{2} Let \tan α = |\frac{Im (z)}{Re (z)}| ∴ \tan α = |\frac{- 1}{1}| = \frac{π}{4} \Rightarrow α = \frac{π}{4} Since point (1, - 1) lies in the fourth quadrant, the argument of z is given by θ = - α = - \frac{π}{4} Polar form = r (\cos θ + i \sin θ) = \sqrt{2} \{\cos (- \frac{π}{4}) + i \sin (- \frac{π}{4})\} = \sqrt{2} (\cos \frac{π}{4} - i \sin \frac{π}{4})$

$(iv) \frac{1 - i}{1 + i} Rationalising the denominator : \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} \Rightarrow \frac{1 + i^{2} - 2 i}{1 - i^{2}} \Rightarrow \frac{- 2 i}{2} (∵ i^{2} = - 1) \Rightarrow - i r = |z| = \sqrt{0 + 1} = 1 Since point (0, - 1) lies on the negative direction of the imaginary axis, the argument of z is given by \frac{3 π}{2} . Polar form = r (\cos θ + i \sin θ) = (c os \frac{3 π}{2} + i \sin \frac{3 π}{2}) = \{c os (2 π - \frac{π}{2}) + i \sin (2 π - \frac{π}{2})\} = (\cos \frac{π}{2} - i \sin \frac{π}{2})$

$(v) \frac{1}{1 + i} Rationalising the denominator : \frac{1}{1 + i} \times \frac{1 - i}{1 - i} \Rightarrow \frac{1 - i}{1 - i^{2}} \Rightarrow \frac{1 - i}{2} (∵ i^{2} = - 1) \Rightarrow \frac{1}{2} - \frac{i}{2} r = |z| = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}} Let \tan α = |\frac{Im (z)}{Re (z)}| ∴ \tan α = |\frac{\frac{1}{2}}{\frac{- 1}{2}}| = 1 \Rightarrow α = \frac{π}{4} Since point (\frac{1}{2}, - \frac{1}{2}) lies in the fourth quadrant, the argument is given by θ = - α = \frac{- π}{4} Polar form = r (\cos θ + i \sin θ) = \frac{1}{\sqrt{2}} \{c os (\frac{- π}{4}) + i \sin (\frac{- π}{4})\} = \frac{1}{\sqrt{2}} (c os \frac{π}{4} - i \sin \frac{π}{4})$

$(v i) \frac{1 + 2 i}{1 - 3 i} Rationalising the denominator : \frac{1 + 2 i}{1 - 3 i} \times \frac{1 + 3 i}{1 + 3 i} \Rightarrow \frac{1 + 3 i + 2 i + 6 i^{2}}{1 - 9 i^{2}} \Rightarrow \frac{- 5 + 5 i}{10} (∵ i^{2} = - 1) \Rightarrow \frac{- 1}{2} + \frac{i}{2} r = |z| = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}} Let \tan α = |\frac{Im (z)}{Re (z)}| Then, \tan α = |\frac{\frac{1}{2}}{\frac{- 1}{2}}| = 1 \Rightarrow α = \frac{π}{4} Since point (\frac{- 1}{2}, \frac{1}{2}) lies in the second quadrant, the argument is given by θ = π - α = π - \frac{π}{4} = \frac{3 π}{4} Polar form = r (\cos θ + i \sin θ) = \frac{1}{\sqrt{2}} (c os \frac{3 π}{4} + i \sin \frac{3 π}{4})$

$(vii) \sin 120 ° - i \cos 120 ° \frac{\sqrt{3}}{2} + \frac{i}{2} r = |z| = \sqrt{\frac{3}{4} + \frac{1}{4}} = 1 Let \tan α = |\frac{Im (z)}{Re (z)}| Then, \tan α = |\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}| = \frac{1}{\sqrt{3}} \Rightarrow α = \frac{π}{6} Since point (\frac{\sqrt{3}}{2}, \frac{1}{2}) lies in the first quadrant, the argument is given by θ = α = \frac{π}{6} Polar form = r (\cos θ + i \sin θ) = \cos \frac{π}{6} + i \sin \frac{π}{6}$

$(viii) \frac{- 16}{1 + i \sqrt{3}} Rationalising the denominator : \frac{- 16}{1 + i \sqrt{3}} \times \frac{1 - i \sqrt{3}}{1 - i \sqrt{3}} \Rightarrow \frac{- 16 + 16 \sqrt{3} i}{1 - 3 i^{2}} \Rightarrow \frac{- 16 + 16 \sqrt{3} i}{4} (∵ i^{2} = - 1) \Rightarrow - 4 + 4 \sqrt{3} i r = |z| = \sqrt{16 + 48} = 8 Let \tan α = |\frac{Im (z)}{Re (z)}| Then, \tan α = |\frac{4 \sqrt{3}}{- 4}| = \sqrt{3} \Rightarrow α = \frac{π}{3} Since the point (- 4, 4 \sqrt{3}) lies in the third quadrant, the argument is given by θ = π - α = π - \frac{π}{3} = \frac{2 π}{3} Polar form = r (\cos θ + i \sin θ) = 8 \{c os \frac{2 π}{3} + i \sin \frac{2 π}{3}\}$

Page No 13.57:

Question 2:

Write (i²⁵)³ in polar form.

Answer:

${(i^{25})}^{3} = i^{75} = i^{4 \times 18 + 3} = {(i^{4})}^{18} . i^{3} = i^{3} [∵ i^{4} = 1] = - i [∵ i^{3} = - i]$

Let $z = 0 - i$ .
Then, $|z| = \sqrt{0^{2} + {(- 1)}^{2}} = 1$ .

Let θ be the argument of z and α be the acute angle given by $\tan α = \frac{|Im (z)|}{|Re (z)|}$ .
Then,
$\tan α = \frac{1}{0} = \infty \Rightarrow α = \frac{π}{2}$

Clearly, z lies in fourth quadrant. So, arg(z) = $- α = - \frac{π}{2}$ .

∴ the polar form of z is $|z| (\cos θ + i \sin θ) = \cos (- \frac{π}{2}) + i \sin (- \frac{π}{2})$ .

Thus, the polar form of (i²⁵)³is $\cos (\frac{π}{2}) - i \sin (\frac{π}{2})$ .

Page No 13.57:

Question 3:

Express the following complex in the form r(cos θ + i sin θ):
(i) 1 + i tan α
(ii) tan α − i
(iii) 1 − sin α + i cos α
(iv) $\frac{1 - i}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$

Answer:

$(i) Let z = 1 + i \tan α ∵ \tan α is periodic with period π . So, let us take α \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π] Case I : When α \in [0, \frac{π}{2}) z = 1 + i \tan α \Rightarrow |z| = \sqrt{1 + \tan^{2} α} = |\sec α| [∵ 0 < α < \frac{π}{2}] = \sec α Let β be an acute angle given by \tan β = |\frac{Im (z)}{Re (z)}| \tan β = |\tan α| = \tan α \Rightarrow β = α As z lies in the first quadrant . Therefore, \arg (z) = β = α Thus, z in the polar form is given by z = \sec α (\cos α + i \sin α) Case II : z = 1 + i \tan α \Rightarrow |z| = \sqrt{1 + \tan^{2} α} = |\sec α| [∵ \frac{π}{2} < α < π] = - \sec α Let β be an acute angle given by \tan β = |\frac{Im (z)}{Re (z)}| \tan β = |\tan α| = - \tan α \Rightarrow \tan β = \tan (π - α) \Rightarrow β = π - α As, z lies in the fourth quadrant . ∴ \arg (z) = - β = α - π Thus, z in the polar form is given by z = - \sec α \{\cos (α - π) + i \sin (α - π)\}$

$(ii) Let z = \tan α - i ∵ \tan α is periodic with period π . So, let us take α \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π] Case I : z = \tan α - i \Rightarrow |z| = \sqrt{\tan^{2} + 1} = |\sec α| [∵ 0 < α < \frac{π}{2}] = \sec α Let β be an acute angle given by \tan β = |\frac{Im (z)}{Re (z)}| \tan β = \frac{1}{|\tan α|} = |\cot α| = \cot α = \tan (\frac{π}{2} - α) \Rightarrow β = \frac{π}{2} - α We can see that Re (z) > 0 and Im (z) < 0 . So, z lies in the fourth quadrant . ∴ \arg (z) = - β = α - \frac{π}{2} Thus, z in the polar form is given by z = \sec α \{\cos (α - \frac{π}{2}) + i \sin (α - \frac{π}{2})\} Case II : z = \tan α - i \Rightarrow |z| = \sqrt{\tan^{2} + 1} = |\sec α| [∵ \frac{π}{2} < α < π] = - \sec α Let β be an acute angle given by \tan β = |\frac{Im (z)}{Re (z)}| \tan β = \frac{1}{|\tan α|} = |\cot α| = - \cot α = \tan (α - \frac{π}{2}) \Rightarrow β = α - \frac{π}{2} We can see that R e (z) < 0 and Im (z) < 0 . So, z lies in the third quadrant . ∴ \arg (z) = π + β = \frac{π}{2} + α Thus, z in the polar form is given by z = - \sec α \{\cos (\frac{π}{2} + α) + i \sin (\frac{π}{2} + α)\}$

$(iii) Let z = (1 - \sin α) + i \cos α . ∵ sine and cosine functions are periodic functions with period 2 π . So, let us take α \in [0, 2 π] Now, z = 1 - \sin α + i \cos α \Rightarrow |z| = \sqrt{{(1 - \sin α)}^{2} + \cos^{2} α} = \sqrt{2 - \sin α} = \sqrt{2} \sqrt{1 - \sin α} \Rightarrow |z| = \sqrt{2} \sqrt{{(\cos \frac{α}{2} - \sin \frac{α}{2})}^{2}} = \sqrt{2} |\cos \frac{α}{2} - \sin \frac{α}{2}| Let β be an acute angle given by \tan β = \frac{|Im (z)|}{|Re (z)|} . Then, \tan β = \frac{|\cos α|}{|1 - \sin α|} = |\frac{\cos^{2} \frac{α}{2} - \sin^{2} \frac{α}{2}}{{(\cos \frac{α}{2} - \sin \frac{α}{2})}^{2}}| = |\frac{\cos \frac{α}{2} + \sin \frac{α}{2}}{\cos \frac{α}{2} - \sin \frac{α}{2}}| \Rightarrow \tan β = |\frac{1 + \tan \frac{α}{2}}{1 - \tan \frac{α}{2}}| = |\tan (\frac{π}{4} + \frac{α}{2})| Case I : When 0 \leq α < \frac{π}{2} In this case, we have, \cos \frac{α}{2} > \sin \frac{α}{2} and \frac{π}{4} + \frac{α}{2} \in [\frac{π}{4}, \frac{π}{2}) \Rightarrow |z| = \sqrt{2} (\cos \frac{α}{2} - \sin \frac{α}{2}) and \tan β = |\tan (\frac{π}{4} + \frac{α}{2})| = \tan (\frac{π}{4} + \frac{α}{2}) \Rightarrow β = \frac{π}{4} + \frac{α}{2} Clearly, z lies in the first quadrant . Therefore, \arg (z) = \frac{π}{4} + \frac{α}{2} Hence, the polar form of z is \sqrt{2} (\cos \frac{α}{2} - \sin \frac{α}{2}) \{\cos (\frac{π}{4} + \frac{α}{2}) + i \sin (\frac{π}{4} + \frac{α}{2})\} Case II : When \frac{π}{2} < α < \frac{3 π}{2} In this case, we have, \cos \frac{α}{2} < \sin \frac{α}{2} and \frac{π}{4} + \frac{α}{2} \in (\frac{π}{2}, π) \Rightarrow |z| = \sqrt{2} (\cos \frac{α}{2} - \sin \frac{α}{2}) = - \sqrt{2} (\cos \frac{α}{2} - \sin \frac{α}{2}) and \tan β = |\tan (\frac{π}{4} + \frac{α}{2})| = - \tan (\frac{π}{4} + \frac{α}{2}) = \tan \{π - (\frac{π}{4} + \frac{α}{2})\} = \tan (\frac{3 π}{4} - \frac{α}{2}) \Rightarrow β = \frac{3 π}{4} - \frac{α}{2} Clearly, z lies in the fourth quadrant . Therefore, \arg (z) = - β = - (\frac{3 π}{4} - \frac{α}{2}) = \frac{α}{2} - \frac{3 π}{4} Hence, the polar form of z is - \sqrt{2} (\cos \frac{α}{2} - \sin \frac{α}{2}) \{\cos (\frac{α}{2} - \frac{3 π}{4}) + i \sin (\frac{α}{2} - \frac{3 π}{4})\} Case III : When \frac{3 π}{2} < α < 2 π In this case, we have, \cos \frac{α}{2} < \sin \frac{α}{2} and \frac{π}{4} + \frac{α}{2} \in (π, \frac{5 π}{4}) \Rightarrow |z| = \sqrt{2} |\cos \frac{α}{2} - \sin \frac{α}{2}| = - \sqrt{2} (\cos \frac{α}{2} - \sin \frac{α}{2}) and \tan β = |\tan (\frac{π}{4} + \frac{α}{2})| = \tan (\frac{π}{4} + \frac{α}{2}) = - \tan \{π - (\frac{π}{4} + \frac{α}{2})\} = \tan (\frac{α}{2} - \frac{3 π}{4}) \Rightarrow β = \frac{α}{2} - \frac{3 π}{4} Clearly, z lies in the first quadrant . Therefore, \arg (z) = β = \frac{α}{2} - \frac{3 π}{4} Hence, the polar form of z is - \sqrt{2} (\cos \frac{α}{2} - \sin \frac{α}{2}) \{\cos (\frac{α}{2} - \frac{3 π}{4}) + i \sin (\frac{α}{2} - \frac{3 π}{4})\}$

$(iv) Let z = \frac{1 - i}{c o s \frac{π}{3} + i s i n \frac{π}{3}} = \frac{1 - i}{\frac{1}{2} + i \frac{\sqrt{3}}{2}} = \frac{2 - 2 i}{1 + i \sqrt{3}} \times \frac{1 - i \sqrt{3}}{1 - i \sqrt{3}} = \frac{2 - 2 i - 2 \sqrt{3} i + 2 \sqrt{3} i^{2}}{1 + 3} = \frac{2 - 2 \sqrt{3} - 2 i (1 + \sqrt{3})}{4} = \frac{(1 - \sqrt{3}) + i (- 1 - \sqrt{3})}{2} = \frac{(1 - \sqrt{3})}{2} + i \frac{(- 1 - \sqrt{3})}{2} Now, z = \frac{(1 - \sqrt{3})}{2} + i \frac{(- 1 - \sqrt{3})}{2} \Rightarrow |z| = \sqrt{{(\frac{1 - \sqrt{3}}{2})}^{2} + {(\frac{- 1 - \sqrt{3}}{2})}^{2}} = \sqrt{(\frac{1 + 3 - 2 \sqrt{3}}{4}) + (\frac{1 + 3 + 2 \sqrt{3}}{4})} = \sqrt{\frac{8}{4}} = \sqrt{2} Let β be an acute angle given by \tan β = \frac{|Im (z)|}{|Re (z)|} . Then, \tan β = \frac{|\frac{1 + \sqrt{3}}{2}|}{|\frac{1 - \sqrt{3}}{2}|} = |\frac{1 + \sqrt{3}}{1 - \sqrt{3}}| = |\frac{\tan \frac{π}{4} + \tan \frac{π}{3}}{1 - \tan \frac{π}{4} \tan \frac{π}{3}}| \Rightarrow \tan β = |\tan (\frac{π}{4} + \frac{π}{3})| = |\tan \frac{7 π}{12}| \Rightarrow β = \frac{7 π}{12} Clearly, z lies in the fourth quadrant . Therefore, \arg (z) = - \frac{7 π}{12} Hence, the polar form of z is \sqrt{2} (\cos \frac{7 π}{12} - \sin \frac{7 π}{12})$

Page No 13.57:

Question 4:

If z₁ and z₂ are two complex numbers such that $|z_{1}| = |z_{2}|$ and arg(z₁) + arg(z₂) = $π$ , then show that $z_{1} = - \bar{z_{2}}$ .

Answer:

Let θ₁be the arg(z₁) and θ₂be the arg(z₂).

It is given that $|z_{1}| = |z_{2}|$ and arg(z₁) + arg(z₂) = $π$ .

Since, z₁ is a complex number.

$z_{1} = |z_{1}| (\cos θ_{1} + i \sin θ_{1}) = |z_{2}| [\cos (π - θ_{2}) + i \sin (π - θ_{2})] = |z_{2}| [- \cos (θ_{2}) + i \sin (θ_{2})] = - |z_{2}| [\cos (θ_{2}) - i \sin (θ_{2})] = - \bar{z_{2}}$

Hence, $z_{1} = - \bar{z_{2}}$ .

Page No 13.57:

Question 5:

If z₁, z₂ and z₃, z₄ are two pairs of conjugate complex numbers, prove that $\arg (\frac{z_{1}}{z_{4}}) + \arg (\frac{z_{2}}{z_{3}}) = 0$ .

Answer:

Given that z₁, z₂ and z₃, z₄ are two pairs of conjugate complex numbers.

$∴ z_{1} = r_{1} e^{i θ_{1}}, z_{2} = r_{1} e^{- i θ_{1}}, z_{3} = r_{2} e^{i θ_{2}} and z_{4} = r_{2} e^{- i θ_{2}}$

Then,
$\frac{z_{1}}{z_{4}} = \frac{r_{1} e^{i θ_{1}}}{r_{2} e^{- i θ_{2}}} = \frac{r_{1}}{r_{2}} e^{i (θ_{1} - θ_{2})} \Rightarrow \arg (\frac{z_{1}}{z_{4}}) = θ_{1} - θ_{2} . . . (1)$
and
$\frac{z_{2}}{z_{3}} = \frac{r_{1} e^{- i θ_{1}}}{r_{2} e^{i θ_{2}}} = \frac{r_{1}}{r_{2}} e^{i (- θ_{1} + θ_{2})} \Rightarrow \arg (\frac{z_{2}}{z_{3}}) = θ_{2} - θ_{1} . . . (2)$

$∴ \arg (\frac{z_{1}}{z_{4}}) + \arg (\frac{z_{2}}{z_{3}}) = θ_{1} - θ_{2} - θ_{1} + θ_{2} = 0$

Hence, $\arg (\frac{z_{1}}{z_{4}}) + \arg (\frac{z_{2}}{z_{3}}) = 0$ .

Page No 13.58:

Question 6:

Express $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$ in polar form.

Answer:

$Let z = \sin \frac{π}{5} + i (1 - \cos \frac{π}{5}) \Rightarrow |z| = \sqrt{{(\sin \frac{π}{5})}^{2} + {(1 - \cos \frac{π}{5})}^{2}} = \sqrt{\sin^{2} \frac{π}{5} + 1 + \cos^{2} \frac{π}{5} - 2 \cos \frac{π}{5}} = \sqrt{2 - 2 \cos \frac{π}{5}} = \sqrt{2} (\sqrt{1 - \cos \frac{π}{5}}) = \sqrt{2} (\sqrt{2 \sin^{2} \frac{π}{10}}) = 2 \sin \frac{π}{10} Let β be an acute angle given by \tan β = \frac{|Im (z)|}{|Re (z)|} . Then, \tan β = \frac{|1 - \cos \frac{π}{5}|}{|\sin \frac{π}{5}|} = |\frac{2 \sin^{2} \frac{π}{10}}{2 \sin \frac{π}{10} \cos \frac{π}{10}}| = |\tan \frac{π}{10}| \Rightarrow β = \frac{π}{10} Clearly, z lies in the first quadrant . Therefore, \arg (z) = \frac{π}{10} Hence, the polar form of z is 2 \sin \frac{π}{10} (\cos \frac{π}{10} + i \sin \frac{π}{10})$

Page No 13.62:

Question 1:

The value of $(1 + i) (1 + i^{2}) (1 + i^{3}) (1 + i^{4})$ is
(a) 2
(b) 0
(c) 1
(d) i

Answer:

(b) 0
(1+ i) (1 + i²) (1 + i³) (1 + i⁴)
= (1+ i) (1 $-$ 1) (1 $-$ i) (1 + 1) ( $∵$ i² = $-$ 1, i³ = $-$ i and i⁴ = 1)
= (1 + i) (0) (1 $-$ i) (2)
= 0

Page No 13.62:

Question 2:

If $\frac{3 + 2 i sinθ}{1 - 2 i sinθ}$ is a real number and 0 < θ < 2π, then θ =
(a) π
(b) $\frac{π}{2}$
(c) $\frac{π}{3}$
(d) $\frac{π}{6}$

Answer:

(a) π

Given:

$\frac{3 + 2 i \sin θ}{1 - 2 i \sin θ}$ is a real number

On rationalising, we get,

$\frac{3 + 2 i \sin θ}{1 - 2 i \sin θ} \times \frac{1 + 2 i \sin θ}{1 + 2 i \sin θ} = \frac{(3 + 2 i \sin θ) (1 + 2 i \sin θ)}{(1)^{2} - (2 i \sin θ)^{2}} = \frac{3 + 2 i \sin θ + 6 i \sin θ + 4 i^{2} \sin^{2} θ}{1 + 4 \sin^{2} θ} = \frac{3 - 4 \sin^{2} θ + 8 i \sin θ}{1 + 4 \sin^{2} θ} [∵ i^{2} = - 1] = \frac{3 - 4 \sin^{2} θ}{1 + 4 \sin^{2} θ} + i \frac{8 \sin θ}{1 + 4 \sin^{2} θ}$

For the above term to be real, the imaginary part has to be zero.

$∴ \frac{8 \sin θ}{1 + 4 \sin^{2} θ} = 0 \Rightarrow 8 \sin θ = 0$

For this to be zero,
sin $θ$ = 0
$\Rightarrow$ $θ$ = 0, $π, 2 π, 3 π . . .$
But $0 < θ < 2 π$
Hence, $θ = π$

Page No 13.62:

Question 3:

If $(1 + i) (1 + 2 i) (1 + 3 i) . . . (1 + n i) = a + i b, then 2 \times 5 \times 10 \times . . . . \times (1 + n^{2})$ is equal to
(a) $\sqrt{a^{2} + b^{2}}$
(b) $\sqrt{a^{2} - b^{2}}$
(c) $a^{2} + b^{2}$
(d) $a^{2} - b^{2}$
(e) $a + b$

Answer:

(c) a²+ b²

(1 + i)(1 + 2i)(1 + 3i) ......(1 + ni) = a + ib

Taking modulus on both the sides, we get:
$|(1 + i) (1 + 2 i) (1 + 3 i) . . . . . . . (1 + n i)| = |a + i b| |(1 + i) (1 + 2 i) (1 + 3 i) . . . . . . . (1 + n i)| can be written as |(1 + i)| |(1 + 2 i)| |(1 + 3 i)| . . . . . . . |(1 + n i)|$

$\sqrt{1^{2} + 1^{2}} \times \sqrt{1^{2} + 2^{2}} \times \sqrt{1^{2} + 3^{2}} \times . . . \times \sqrt{1 + n^{2}} = \sqrt{a^{2} + b^{2}}$

$\Rightarrow \sqrt{2} \times \sqrt{5} \times \sqrt{10} \times . . . \times \sqrt{1 + n^{2}} = \sqrt{a^{2} + b^{2}}$

Squaring on both the sides, we get:

$2 \times 5 \times 10 \times . . . \times (1 + n^{2}) = a^{2} + b^{2}$
2

Page No 13.62:

Question 4:

If $\sqrt{a + i b} = x + i y,$ then possible value of $\sqrt{a - i b}$ is
(a) $x^{2} + y^{2}$
(b) $\sqrt{x^{2} + y^{2}}$
(c) x + iy
(d) x − iy
(e) $\sqrt{x^{2} - y^{2}}$

Answer:

(d) x $-$ iy

$\sqrt{a + i b} = x + i y Squaring on both the sides, we get, a + i b = x^{2} + (i y)^{2} + 2 i x y \Rightarrow a + i b = (x^{2} - y^{2}) + 2 i x y ∴ a = (x^{2} - y^{2}) and b = 2 x y ∴ a - i b = (x^{2} - y^{2}) - 2 i x y \Rightarrow a - i b = x^{2} + i^{2} y^{2} - 2 i x y [∵ i^{2} = - 1] Taking square root on both the sides, we get : \sqrt{a - i b} = x - i y$

Page No 13.62:

Question 5:

If $z = \cos \frac{π}{4} + i \sin \frac{π}{6}$ , then
(a) $|z| = 1, a r g (z) = \frac{π}{4}$
(b) $|z| = 1, a r g (z) = \frac{π}{6}$
(c) $|z| = \frac{\sqrt{3}}{2}, a r g (z) = \frac{5 π}{24}$
(d) $|z| = \frac{\sqrt{3}}{2}, a r g (z) = \tan^{- 1} \frac{1}{\sqrt{2}}$

Answer:

(d) $|z| = \frac{\sqrt{3}}{2}, \arg (z) = \tan^{- 1} \frac{1}{\sqrt{2}}$
$z = \cos \frac{π}{4} + i \sin \frac{π}{6} \Rightarrow z = \frac{1}{\sqrt{2}} + \frac{1}{2} i \Rightarrow |z| = \sqrt{{(\frac{1}{\sqrt{2}})}^{2} + \frac{1}{4}} \Rightarrow |z| = \sqrt{\frac{1}{2} + \frac{1}{4}} \Rightarrow |z| = \sqrt{\frac{3}{4}} \Rightarrow |z| = \frac{\sqrt{3}}{2} \tan α = |\frac{Im (z)}{Re (z)}| = \frac{1}{\sqrt{2}} \Rightarrow α = \tan^{- 1} (\frac{1}{\sqrt{2}}) Since, the point z lies in the first quadrant . Therefore, \arg (z) = α = \tan^{- 1} (\frac{1}{\sqrt{2}})$

Page No 13.62:

Question 6:

The polar form of (i²⁵)³ is
(a) $\cos \frac{π}{2} + i \sin \frac{π}{2}$
(b) cos π + i sin π
(c) cos π − i sin π
(d) $\cos \frac{π}{2} - i \sin \frac{π}{2}$

Answer:

(d) $\cos \frac{π}{2} - i \sin \frac{π}{2}$
(i²⁵)³ = (i)⁷⁵
         = (i)^{4 $\times$ 18+ 3}
         = (i)³
         = $-$ i           ( $∵$ i⁴= 1)

$Let z = 0 - i Since, the point (0, - 1) lies on the negative direction of imaginary axis . Therefore, \arg (z) = \frac{- π}{2}$

Modulus, r = $|z| = |1| = 1$

$∴$ Polar form = r (cos $θ$ + i sin $θ$ )
                      = cos $(\frac{- π}{2})$ +i sin $(\frac{- π}{2})$
                      = cos $\frac{π}{2}$ $-$ i sin $\frac{π}{2}$

Page No 13.62:

Question 7:

If i² = −1, then the sum i + i² + i³ +... upto 1000 terms is equal to
(a) 1
(b) −1
(c) i
(d) 0

Answer:

(d) 0

$i + i^{2} + i^{3} + i^{4} . . . i^{1000} i + i^{2} + i^{3} + i^{4} [∵ i^{2} = - 1, i^{3} = - i and i^{4} = 1] = i - 1 - i + 1 = 0 Similarly, the sum of the next four terms of the series will be equal to 0 . This is because the powers of i follow a cyclicity of 4 . Hence, the sum of all terms, till 1000, will be zero . i + i^{2} + i^{3} + i^{4} . . . i^{1000} = 0$

Page No 13.63:

Question 8:

If $z = \frac{- 2}{1 + i \sqrt{3}}$ , then the value of arg (z) is
(a) π
(b) $\frac{π}{3}$
(c) $\frac{2 π}{3}$
(d) $\frac{π}{4}$

Answer:

(c) $\frac{2 π}{3}$
z = $\frac{- 2}{1 + i \sqrt{3}}$

Rationalising z, we get,

$z = \frac{- 2}{1 + i \sqrt{3}} \times \frac{1 - i \sqrt{3}}{1 - i \sqrt{3}} \Rightarrow z = \frac{- 2 + i 2 \sqrt{3}}{1 + 3} \Rightarrow z = \frac{- 1 + i \sqrt{3}}{2} \Rightarrow z = \frac{- 1}{2} + \frac{i \sqrt{3}}{2}$
$\tan α = |\frac{Im (z)}{Re (z)}| = \sqrt{3} \Rightarrow α = \frac{π}{3} Since, z lies in the second quadrant . Therefore, \arg (z) = π - \frac{π}{3} = \frac{2 π}{3}$

Page No 13.63:

Question 9:

If a = cos θ + i sin θ, then $\frac{1 + a}{1 - a} =$
(a) $\cot \frac{θ}{2}$
(b) cot θ
(c) $i \cot \frac{θ}{2}$
(d) $i \tan \frac{θ}{2}$

Answer:

(c) $i c o t \frac{θ}{2}$
$a = \cos θ + i \sin θ (given) \Rightarrow \frac{1 + a}{1 - a} = \frac{1 + \cos θ + i \sin θ}{1 - \cos θ - i \sin θ} \Rightarrow \frac{1 + a}{1 - a} = \frac{1 + \cos θ + i \sin θ}{1 - \cos θ - i \sin θ} \times \frac{1 - \cos θ + i \sin θ}{1 - \cos θ + i \sin θ}$

$\Rightarrow \frac{1 + a}{1 - a} = \frac{{(1 + i \sin θ)}^{2} - \cos^{2} θ}{{(1 - \cos θ)}^{2} - {(i \sin θ)}^{2}}$

$\Rightarrow \frac{1 + a}{1 - a} = \frac{1 - \sin^{2} θ + 2 i \sin θ - \cos^{2} θ}{1 + \cos^{2} θ - 2 \cos θ + \sin^{2} θ}$

$\Rightarrow \frac{1 + a}{1 - a} = \frac{1 - (\sin^{2} θ + \cos^{2} θ) + 2 i \sin θ}{1 + (\sin^{2} θ + \cos^{2} θ) - 2 \cos θ}$

$\Rightarrow \frac{1 + a}{1 - a} = \frac{2 i \sin θ}{2 (1 - \cos θ)}$
$\Rightarrow \frac{1 + a}{1 - a} = \frac{2 i \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}}$
$\Rightarrow \frac{1 + a}{1 - a} = \frac{i \cos \frac{θ}{2}}{\sin \frac{θ}{2}}$
$\Rightarrow \frac{1 + a}{1 - a} = i \cot \frac{θ}{2}$

Page No 13.63:

Question 10:

If (1 + i) (1 + 2i) (1 + 3i) .... (1 + ni) = a + ib, then 2.5.10.17.......(1+n²)=
(a) a − ib
(b) a² − b²
(c) a² + b²
(d) none of these

Answer:

(c) a²+ b²

(1 + i)(1 + 2i)(1 + 3i) ......(1 + ni) = a + ib

Taking modulus on both the sides, we get,

$|(1 + i) (1 + 2 i) (1 + 3 i) . . . . . . (1 + n i)| = a + i b |(1 + i) (1 + 2 i) (1 + 3 i) . . . . . . (1 + n i)| can be wriiten as |(1 + i)| |(1 + 2 i)| |(1 + 3 i)| . . . . |(1 + n i)| ∴ \sqrt{1^{2} + 1^{2}} \times \sqrt{1^{2} + 2^{2}} \times \sqrt{1^{2} + 3^{2}} . . . . \times \sqrt{1^{2} + n^{2}} = \sqrt{a^{2} + b^{2}}$

$\Rightarrow \sqrt{2} \times \sqrt{5} \times \sqrt{10} . . . . \times \sqrt{1 + n^{2}} = \sqrt{a^{2} + b^{2}}$

Squaring on both the sides, we get:

$2 \times 5 \times 10 . . . . \times (1 + n^{2}) = a^{2} + b^{2}$
2×5×10×.....(1 + n2) = a2 + b2

Page No 13.63:

Question 11:

If $\frac{(a^{2} + 1)^{2}}{2 a - i} = x + i y, then x^{2} + y^{2}$ is equal to
(a) $\frac{(a^{2} + 1)^{4}}{4 a^{2} + 1}$
(b) $\frac{(a + 1)^{2}}{4 a^{2} + 1}$
(c) $\frac{(a^{2} - 1)^{2}}{(4 a^{2} - 1)^{2}}$
(d) none of these

Answer:

(a) $\frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1}$
$x + i y = \frac{{(a^{2} + 1)}^{2}}{2 a - i}$

Taking modulus on both the sides, we get:
$\sqrt{x^{2} + y^{2}} = \frac{{(a^{2} + 1)}^{2}}{\sqrt{4 a^{2} + 1}}$
$Squaring both sides, we get, x^{2} + y^{2} = \frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1}$

Page No 13.63:

Question 12:

The principal value of the amplitude of (1 + i) is
(a) $\frac{π}{4}$
(b) $\frac{π}{12}$
(c) $\frac{3 π}{4}$
(d) π

Answer:

(a) $\frac{π}{4}$

Let z = (1+i)
$\tan α = |\frac{Im (z)}{Re (z)}| = 1 \Rightarrow α = \frac{π}{4} Since, z lies in the first quadrant .$
Therefore, arg (z) = $\frac{π}{4}$

Page No 13.63:

Question 13:

The least positive integer n such that ${(\frac{2 i}{1 + i})}^{n}$ is a positive integer, is
(a) 16
(b) 8
(c) 4
(d) 2

Answer:

$(b) 8 Let z = (\frac{2 i}{1 + i}) \Rightarrow z = \frac{2 i}{1 + i} \times \frac{1 - i}{1 - i} \Rightarrow z = \frac{2 i (1 - i)}{1 - i^{2}} \Rightarrow z = \frac{2 i (1 - i)}{1 + 1} [∵ i^{2} = - 1] \Rightarrow z = \frac{2 i (1 - i)}{2} \Rightarrow z = i - i^{2} \Rightarrow z = i + 1 Now, z^{n} = {(1 + i)}^{n} For n = 2, z^{2} = {(1 + i)}^{2} = 1 + i^{2} + 2 i = 1 - 1 + 2 i = 2 i . . . (1) Since this is not a positive integer, For n = 4, z^{4} = {(1 + i)}^{4} = {[{(1 + i)}^{2}]}^{2} = {(2 i)}^{2} [Using (1)] = 4 i^{2} = - 4 . . . (2) This is a negative integer . For n = 8, z^{8} = {(1 + i)}^{8} = {[{(1 + i)}^{4}]}^{2} = {(- 4)}^{2} [Using (2)] = 16 This is a positive integer . Thus, z = {(\frac{2 i}{1 + i})}^{n} is positive for n = 8 . Therefore, 8 is the least positive integer such that {(\frac{2 i}{1 + i})}^{n} is a positive integer .$

Page No 13.63:

Question 14:

If z is a non-zero complex number, then $|\frac{{|z|}^{2}}{z z}|$ is equal to
(a) $|\frac{z}{z}|$
(b) $|z|$
(c) $|z|$
(d) none of these

Answer:

(a) $|\frac{z}{z}|$
$|\frac{{|z|}^{2}}{z z}| = |\frac{{|z|}^{2}}{{|z|}^{2}}| (∵ z z = {|z|}^{2}) Let z = a + i b \Rightarrow |z| = \sqrt{a^{2} + b^{2}} Let z = a - i b \Rightarrow |z| = \sqrt{a^{2} + b^{2}} ∴ |\frac{{|z|}^{2}}{z z}| = |\frac{{|z|}^{2}}{{|z|}^{2}}| = |\frac{z}{z}|$

Page No 13.63:

Question 15:

If a = 1 + i, then a² equals
(a) 1 − i
(b) 2i
(c) (1 + i) (1 − i)
(d) i − 1.

Answer:

(b) 2i

a = 1 + i
On squaring both the sides, we get,
a² = (1 + i)²
$\Rightarrow$ a² = 1 + i² + 2i
$\Rightarrow$ a² = 1 $-$ 1 + 2i ( $∵$ i² = $-$ 1)
$\Rightarrow$ a² = 2i

Page No 13.63:

Question 16:

If (x + iy)1/3 = a + ib, then $\frac{x}{a} + \frac{y}{b} =$
(a) 0
(b) 1
(c) −1
(d) none of these

Answer:

(d) none of these

${(x + i y)}^{\frac{1}{3}} = a + i b Cubing on both the sides, we get : x + i y = {(a + i b)}^{3} \Rightarrow x + i y = a^{3} + {(i b)}^{3} + 3 a^{2} b i + 3 a {(i b)}^{2} \Rightarrow x + i y = a^{3} + i^{3} b^{3} + 3 a^{2} i b + 3 i^{2} a b^{2} \Rightarrow x + i y = a^{3} - i b^{3} + 3 a^{2} i b - 3 a b^{2} (∵ i^{2} = - 1, i^{3} = - i) \Rightarrow x + i y = a^{3} - 3 a b^{2} + i (- b^{3} + 3 a^{2} b) ∴ x = a^{3} - 3 a b^{2} and y = 3 a^{2} b - b^{3} or, \frac{x}{a} = a^{2} - 3 b^{2} and \frac{y}{b} = 3 a^{2} - b^{2} \Rightarrow \frac{x}{a} + \frac{y}{b} = a^{2} - 3 b^{2} + 3 a^{2} - b^{2} \Rightarrow \frac{x}{a} + \frac{y}{b} = 4 a^{2} - 4 b^{2}$

Page No 13.63:

Question 17:

$(\sqrt{- 2}) (\sqrt{- 3})$ is equal to
(a) $\sqrt{6}$
(b) $- \sqrt{6}$
(c) $i \sqrt{6}$
(d) none of these.

Answer:

(b) $- \sqrt{6}$

$\sqrt{- 2} \times \sqrt{- 3} = \sqrt{2} \times \sqrt{3} \times \sqrt{- 1} \times \sqrt{- 1} = \sqrt{6} \times i \times i = \sqrt{6} \times i^{2} = - \sqrt{6} [∵ i^{2} = - 1]$

Page No 13.63:

Question 18:

The argument of $\frac{1 - i \sqrt{3}}{1 + i \sqrt{3}}$ is
(a) 60°
(b) 120°
(c) 210°
(d) 240°

Answer:

(d) 240°

$\frac{1 - i \sqrt{3}}{1 + i \sqrt{3}} Rationalising the denominator, \frac{1 - i \sqrt{3}}{1 + i \sqrt{3}} \times \frac{1 - i \sqrt{3}}{1 - i \sqrt{3}} = \frac{1 + 3 i^{2} - 2 \sqrt{3} i}{1 - 3 i^{2}} = \frac{- 2 - 2 \sqrt{3} i}{4} (∵ i^{2} = - 1) = \frac{- 1}{2} - i \frac{\sqrt{3}}{2} \tan α = |\frac{Im (z)}{Re (z)}| Then, \tan α = |\frac{\frac{- \sqrt{3}}{2}}{\frac{- 1}{2}}| = \sqrt{3} \Rightarrow α = 60 ° Since the points (\frac{- 1}{2}, \frac{- \sqrt{3}}{2}) lie in the third quadrant, the argument is given by : θ = 180 ° + 60 ° = 240 °$

Page No 13.63:

Question 19:

If $z = (\frac{1 + i}{1 - i})$ , then z⁴ equals
(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

(a) 1
$Let z = \frac{1 + i}{1 - i}$ ^{Rationalising the denominator:
$z = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}$
$\Rightarrow z = \frac{1 + i^{2} + 2 i}{1 - i^{2}}$
$\Rightarrow z = \frac{1 - 1 + 2 i}{1 + 1}$
$\Rightarrow z = \frac{2 i}{2} \Rightarrow z = i$
$\Rightarrow z^{4} = i^{4} Since i^{2} = - 1, we have : \Rightarrow z^{4} = i^{2} \times i^{2} \Rightarrow z^{4} = 1$}

Page No 13.63:

Question 20:

If $z = \frac{1 + 2 i}{1 - (1 - i)^{2}}$ , then arg (z) equal
(a) 0
(b) $\frac{π}{2}$
(c) π
(d) none of these.

Answer:

(a) 0

$Let z = \frac{1 + 2 i}{1 - {(1 - i)}^{2}}$
$\Rightarrow z = \frac{1 + 2 i}{1 - (1 + i^{2} - 2 i)}$
$\Rightarrow z = \frac{1 + 2 i}{1 - (1 - 1 - 2 i)}$
$\Rightarrow z = \frac{1 + 2 i}{1 + 2 i}$
$\Rightarrow z = 1 Since point (1, 0) lies on the positive direction of real axis, we have : \arg (z) = 0$

Page No 13.64:

Question 21:

$If z = \frac{1}{(2 + 3 i)^{2}}, than |z| =$
(a) $\frac{1}{13}$
(b) $\frac{1}{5}$
(c) $\frac{1}{12}$
(d) none of these

Answer:

(a) $\frac{1}{13}$
$Let z = \frac{1}{{(2 + 3 i)}^{2}} \Rightarrow z = \frac{1}{4 + 9 i^{2} + 12 i} \Rightarrow z = \frac{1}{4 - 9 + 12 i} \Rightarrow z = \frac{1}{- 5 + 12 i}$

$\Rightarrow z = \frac{1}{- 5 + 12 i} \times \frac{- 5 - 12 i}{- 5 - 12 i}$

$\Rightarrow z = \frac{- 5 - 12 i}{25 + 144} \Rightarrow z = \frac{- 5}{169} - \frac{12 i}{169}$

$\Rightarrow |z| = \sqrt{\frac{25}{169^{2}} + \frac{144}{169^{2}}}$
$\Rightarrow |z| = \frac{1}{\sqrt{169}}$
$\Rightarrow |z| = \frac{1}{13}$

Page No 13.64:

Question 22:

$If z = \frac{1}{(1 - i) (2 + 3 i)}, than |z| =$
(a) 1
(b) $1 / \sqrt{26}$
(c) $5 / \sqrt{26}$
(d) none of these

Answer:

(b) $\frac{1}{\sqrt{26}}$
$Let z = \frac{1}{(1 - i) (2 + 3 i)} \Rightarrow z = \frac{1}{2 + i - 3 i^{2}} \Rightarrow z = \frac{1}{2 + i + 3}$

$\Rightarrow z = \frac{1}{5 + i} \times \frac{5 - i}{5 - i}$

$\Rightarrow z = \frac{5 - i}{25 - i^{2}}$

$\Rightarrow z = \frac{5 - i}{25 + 1}$

$\Rightarrow z = \frac{5 - i}{26}$

$\Rightarrow z = \frac{5}{26} - \frac{i}{26}$

$\Rightarrow |z| = \sqrt{\frac{25}{676} + \frac{1}{676}}$
$\Rightarrow z = \frac{1}{\sqrt{26}}$

Page No 13.64:

Question 23:

$If z = 1 - cosθ + i sinθ, then |z| =$
(a) $2 \sin \frac{θ}{2}$
(b) $2 \cos \frac{θ}{2}$
(c) $2 |\sin \frac{θ}{2}|$
(d) $2 |\cos \frac{θ}{2}|$

Answer:

(c) $2 |\sin \frac{θ}{2}|$

$∵ z = 1 - \cos θ + i \sin θ \Rightarrow |z| = \sqrt{{(1 - \cos θ)}^{2} + \sin^{2} θ} \Rightarrow |z| = \sqrt{1 + \cos^{2} θ - 2 \cos θ + \sin^{2} θ} \Rightarrow |z| = \sqrt{1 + 1 - 2 \cos θ} \Rightarrow |z| = \sqrt{2 (1 - \cos θ)} \Rightarrow |z| = \sqrt{4 \sin^{2} \frac{θ}{2}} \Rightarrow |z| = 2 |\sin \frac{θ}{2}|$

Page No 13.64:

Question 24:

If $x + i y = (1 + i) (1 + 2 i) (1 + 3 i)$ , then x² + y² =
(a) 0
(b) 1
(c) 100
(d) none of these

Answer:

(c) 100

$∵ x + i y = (1 + i) (1 + 2 i) (1 + 3 i) Taking modulus on both the sides : |x + i y| = |(1 + i) (1 + 2 i) (1 + 3 i)| \Rightarrow |x + i y| = |1 + i| \times |1 + 2 i| \times |1 + 3 i| \Rightarrow \sqrt{x^{2} + y^{2}} = \sqrt{1^{2} + 1^{2}} \sqrt{1^{2} + 2^{2}} \sqrt{1^{2} + 3^{2}} \Rightarrow \sqrt{x^{2} + y^{2}} = \sqrt{2} \sqrt{5} \sqrt{10} \Rightarrow \sqrt{x^{2} + y^{2}} = \sqrt{100} Squaring both the sides, \Rightarrow x^{2} + y^{2} = 100$

Page No 13.64:

Question 25:

If $z = \frac{1}{1 - cosθ - i sinθ}$ , then Re (z) =
(a) 0
(b) $\frac{1}{2}$
(c) $\cot \frac{θ}{2}$
(d) $\frac{1}{2} \cot \frac{θ}{2}$

Answer:

(b) $\frac{1}{2}$
$z = \frac{1}{1 - \cos θ - i \sin θ} z = \frac{1}{1 - \cos θ - i \sin θ} \times \frac{1 - \cos θ + i \sin θ}{1 - \cos θ + i \sin θ} \Rightarrow z = \frac{1 - \cos θ + i \sin θ}{{(1 - \cos θ)}^{2} - {(i \sin θ)}^{2}} \Rightarrow z = \frac{1 - \cos θ + i \sin θ}{1 + \cos^{2} θ - 2 \cos θ + \sin^{2} θ} \Rightarrow z = \frac{1 - \cos θ + i \sin θ}{1 + 1 - 2 \cos θ} \Rightarrow z = \frac{1 - \cos θ + i \sin θ}{2 (1 - \cos θ)} \Rightarrow R e (z) = \frac{(1 - \cos θ)}{2 (1 - \cos θ)} = \frac{1}{2}$

Page No 13.64:

Question 26:

If $x + i y = \frac{3 + 5 i}{7 - 6 i},$ then y =
(a) 9/85
(b) −9/85
(c) 53/85
(d) none of these

Answer:

(c) $\frac{53}{85}$
$x + i y = \frac{3 + 5 i}{7 - 6 i} \Rightarrow x + i y = \frac{3 + 5 i}{7 - 6 i} \times \frac{7 + 6 i}{7 + 6 i} \Rightarrow x + i y = \frac{21 + 53 i + 30 i^{2}}{49 - 36 i^{2}} \Rightarrow x + i y = \frac{21 - 30 + 53 i}{49 + 36} \Rightarrow x + i y = \frac{- 9}{85} + i \frac{53}{85} O n c o m p a r i n g b o t h the s i d e s : y = \frac{53}{85}$

Page No 13.64:

Question 27:

If $\frac{1 - i x}{1 + i x} = a + i b$ , then $a^{2} + b^{2}$ =
(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

(a) 1

$\frac{1 - i x}{1 + i x} = a + i b Taking modulus on both the sides, we get : |\frac{1 - i x}{1 + i x}| = |a + i b| \Rightarrow \frac{\sqrt{1^{2} + x^{2}}}{\sqrt{1^{2} + x^{2}}} = \sqrt{a^{2} + b^{2}} \Rightarrow \sqrt{a^{2} + b^{2}} = 1 Squaring both the sides, we get : a^{2} + b^{2} = 1$

Page No 13.64:

Question 28:

If θ is the amplitude of $\frac{a + i b}{a - i b}$ , than tan θ =
(a) $\frac{2 a}{a^{2} + b^{2}}$
(b) $\frac{2 a b}{a^{2} - b^{2}}$
(c) $\frac{a^{2} - b^{2}}{a^{2} + b^{2}}$
(d) none of these

Answer:

(b) $\frac{2 a b}{a^{2} - b^{2}}$

$z = \frac{a + i b}{a - i b} \times \frac{a + i b}{a + i b} \Rightarrow z = \frac{a^{2} + i^{2} b^{2} + 2 a b i}{a^{2} - i^{2} b^{2}} \Rightarrow z = \frac{a^{2} - b^{2} + 2 a b i}{a^{2} + b^{2}} \Rightarrow z = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} + i \frac{2 a b}{a^{2} + b^{2}} \Rightarrow Re (z) = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}, Im (z) = \frac{2 a b}{a^{2} + b^{2}} \tan α = |\frac{Im (z)}{Re (z)}| = \frac{2 a b}{a^{2} - b^{2}} α = \tan^{- 1} (\frac{2 a b}{a^{2} - b^{2}}) Since, z lies in the first quadrant . Therefore, \arg (z) = α = \tan^{- 1} (\frac{2 a b}{a^{2} - b^{2}}) \tan θ = \frac{2 a b}{a^{2} - b^{2}}$

Page No 13.64:

Question 29:

If $z = \frac{1 + 7 i}{(2 - i)^{2}}$ , then
(a) $|z| = 2$
(b) $|z| = \frac{1}{2}$
(c) amp (z) = $\frac{π}{4}$
(d) amp (z) = $\frac{3 π}{4}$

Answer:

(d) amp (z) = $\frac{3 π}{4}$

$z = \frac{1 + 7 i}{{(2 - i)}^{2}} \Rightarrow z = \frac{1 + 7 i}{4 + i^{2} - 4 i} \Rightarrow z = \frac{1 + 7 i}{4 - 1 - 4 i} [∵ i^{2} = - 1] \Rightarrow z = \frac{1 + 7 i}{3 - 4 i} \Rightarrow z = \frac{1 + 7 i}{3 - 4 i} \times \frac{3 + 4 i}{3 + 4 i} \Rightarrow z = \frac{3 + 4 i + 21 i + 28 i^{2}}{9 - 16 i^{2}} \Rightarrow z = \frac{3 - 28 + 25 i}{9 + 16} \Rightarrow z = \frac{- 25 + 25 i}{25} \Rightarrow z = - 1 + i \tan α = |\frac{Im (z)}{Re (z)}| = 1 \Rightarrow α = \frac{π}{4} Since, z lies in the second quadrant . Therefore, amp (z) = π - α = π - \frac{π}{4} = \frac{3 π}{4}$

Page No 13.64:

Question 30:

The amplitude of $\frac{1}{i}$ is equal to
(a) 0
(b) $\frac{π}{2}$
(c) $- \frac{π}{2}$
(d) π

Answer:

(c) $- \frac{π}{2}$

$Let z = \frac{1}{i} \Rightarrow z = \frac{1}{i} \times \frac{i}{i} \Rightarrow z = \frac{i}{i^{2}} \Rightarrow z = - i$

$Since, z (0, - 1) lies on the negative imaginary axis . Therefore, \arg (z) = \frac{- π}{2}$

Page No 13.64:

Question 31:

The argument of $\frac{1 - i}{1 + i}$ is
(a) $- \frac{π}{2}$
(b) $\frac{π}{2}$
(c) $\frac{3 π}{2}$
(d) $\frac{5 π}{2}$

Answer:

(a) $- \frac{π}{2}$

$Let z = \frac{1 - i}{1 + i} \Rightarrow z = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} \Rightarrow z = \frac{1 + i^{2} - 2 i}{1 - i^{2}} \Rightarrow z = \frac{1 - 1 - 2 i}{1 + 1} \Rightarrow z = \frac{- 2 i}{2} \Rightarrow z = - i Since, z lies on negative direction of imaginary axis . Therefore, \arg (z) = \frac{- π}{2}$

Page No 13.64:

Question 32:

The amplitude of $\frac{1 + i \sqrt{3}}{\sqrt{3} + i}$ is
(a) $\frac{π}{3}$
(b) $- \frac{π}{3}$
(c) $\frac{π}{6}$
(d) $- \frac{π}{6}$

Answer:

(c) $\frac{π}{6}$

$Let z = \frac{1 + i \sqrt{3}}{\sqrt{3} + i} \Rightarrow z = \frac{1 + i \sqrt{3}}{\sqrt{3} + i} \times \frac{\sqrt{3} - i}{\sqrt{3} - i} \Rightarrow z = \frac{\sqrt{3} + 2 i - \sqrt{3} i^{2}}{3 - i^{2}} \Rightarrow z = \frac{\sqrt{3} + \sqrt{3} + 2 i}{4} \Rightarrow z = \frac{2 \sqrt{3} + 2 i}{4} \Rightarrow z = \frac{\sqrt{3}}{2} + \frac{1}{2} i \tan α = |\frac{Im (z)}{Re (z)}| = \frac{1}{\sqrt{3}} \Rightarrow α = \frac{π}{6} Since, z lies in the first quadrant . Therefore, a r g (z) = \tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6}$

Page No 13.65:

Question 33:

The value of (i⁵ + i⁶ + i⁷ + i⁸ + i⁹) / (1 + i) is
(a) $\frac{1}{2} (1 + i)$
(b) $\frac{1}{2} (1 - i)$
(c) 1
(d) $\frac{1}{2}$

Answer:

(a) $\frac{1}{2} (1 + i)$

$\frac{i^{5} + i^{6} + i^{7} + i^{8} + i^{9}}{1 + i} = \frac{i - 1 - i + 1 + i}{1 + i} [As, i^{5} = i, i^{6} = - 1, i^{7} = - i, i^{8} = 1, i^{9} = i] = \frac{i}{i + 1} = \frac{i}{i + 1} \times \frac{i - 1}{i - 1} = \frac{i (i - 1)}{i^{2} - 1} = \frac{i^{2} - i}{- 2} = \frac{1}{2} (1 + i)$

Page No 13.65:

Question 34:

$\frac{1 + 2 i + 3 i^{2}}{1 - 2 i + 3 i^{2}}$ equals
(a) i
(b) −1
(c) −i
(d) 4

Answer:

(c) $-$ i

$Let z = \frac{1 + 2 i + 3 i^{2}}{1 - 2 i + 3 i^{2}} \Rightarrow z = \frac{1 + 2 i - 3}{1 - 2 i - 3} \Rightarrow z = \frac{- 2 + 2 i}{- 2 - 2 i} \times \frac{- 2 + 2 i}{- 2 + 2 i} \Rightarrow z = \frac{{(- 2 + 2 i)}^{2}}{{(- 2)}^{2} - {(2 i)}^{2}} \Rightarrow z = \frac{4 + 4 i^{2} - 8 i}{4 + 4} \Rightarrow z = \frac{4 - 4 - 8 i}{8} \Rightarrow z = \frac{- 8 i}{8} \Rightarrow z = - i$

Page No 13.65:

Question 35:

The value of $\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} - 1$ is
(a) −1
(b) −2
(c) −3
(d) −4

Answer:

(b) $-$ 2

$\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} - 1 = \frac{i^{4 \times 148} + i^{4 \times 147 + 2} + i^{4 \times 147} + i^{4 \times 146 + 2} + i^{4 \times 146}}{i^{4 \times 145 + 2} + i^{4 \times 145} + i^{4 \times 144 + 2} + i^{4 \times 144} + i^{4 \times 143 + 2}} - 1 [∵ i^{4} = 1 and i^{2} = - 1] = \frac{1 + i^{2} + 1 + i^{2} + 1}{i^{2} + 1 + i^{2} + 1 + i^{2}} - 1 = \frac{1}{- 1} - 1 = - 2$

Page No 13.65:

Question 36:

The value of $(1 + i)^{4} + (1 - i)^{4}$ is
(a) 8
(b) 4
(c) −8
(d) −4

Answer:

(c) $-$ 8

$Using a^{4} + b^{4} = {(a^{2} + b^{2})}^{2} - 2 a^{2} b^{2} (1 + i)^{4} + (1 - i)^{4} = {({(1 + i)}^{2} + {(1 - i)}^{2})}^{2} - 2 {(1 + i)}^{2} {(1 - i)}^{2} = {(1 + i^{2} + 2 i + 1 + i^{2} - 2 i)}^{2} - 2 (1 + i^{2} + 2 i) (1 + i^{2} - 2 i) = {(1 - 1 + 2 i + 1 - 1 - 2 i)}^{2} - 2 (1 - 1 + 2 i) (1 - 1 - 2 i) = (0) - 2 (2 i) (- 2 i) (∵ i^{2} = - 1) = 8 i^{2} = - 8$

Page No 13.65:

Question 37:

If $z = a + i b$ lies in third quadrant, then $\frac{\bar{z}}{z}$ also lies in third quadrant if

(a) $a > b > 0$
(b) $a < b < 0$
(c) $b < a < 0$
(d) $b > a > 0$

Answer:

Since, $z = a + i b$ lies in third quadrant.
$\Rightarrow a < 0 and b < 0 . . . . (1)$

Now,
$\frac{\bar{z}}{z} = \frac{\bar{a + i b}}{a + i b} = \frac{a - i b}{a + i b} = \frac{a - i b}{a + i b} \times \frac{a - i b}{a - i b} = \frac{a^{2} + i^{2} b^{2} - 2 a b i}{a^{2} - i^{2} b^{2}} = \frac{a^{2} - b^{2} - 2 a b i}{a^{2} + b^{2}}$

Since, $\frac{\bar{z}}{z}$ also lies in third quadrant.
$\Rightarrow a^{2} - b^{2} < 0 \Rightarrow (a - b) (a + b) < 0 \Rightarrow a - b > 0 and a + b < 0 \Rightarrow a > b . . . . (2)$

From (1) and (2),
$b < a < 0$

Hence, the correct option is (c).

Page No 13.65:

Question 38:

If $f (z) = \frac{7 - z}{1 - z^{2}}$ , where $z = 1 + 2 i$ , then $|f (z)|$ is

(a) $\frac{|z|}{2}$
(b) $|z|$
(c) $2 |z|$
(d) none of these

Answer:

$f (z) = \frac{7 - z}{1 - z^{2}} = \frac{7 - (1 + 2 i)}{1 - {(1 + 2 i)}^{2}} = \frac{7 - 1 - 2 i}{1 - (1^{2} + 2^{2} i^{2} + 4 i)} = \frac{6 - 2 i}{1 - 1 + 4 - 4 i} = \frac{6 - 2 i}{4 - 4 i} = \frac{6 - 2 i}{4 - 4 i} \times \frac{4 + 4 i}{4 + 4 i} = \frac{24 + 24 i - 8 i - 8 i^{2}}{4^{2} - 4^{2} i^{2}} = \frac{24 + 16 i + 8}{16 + 16} = \frac{32 + 16 i}{32} = 1 + \frac{1}{2} i$

Since $z = 1 + 2 i$ ,
$∴ |z| = \sqrt{{(1)}^{2} + {(2)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$

$∴ |f (z)| = \sqrt{{(1)}^{2} + {(\frac{1}{2})}^{2}} = \sqrt{1 + \frac{1}{4}} = \frac{\sqrt{5}}{2} = \frac{|z|}{2}$

Hence, the correct answer is option (a).

Page No 13.65:

Question 39:

A real value of x satisfies the equation $\frac{3 - 4 i x}{3 + 4 i x} = a - i b (a, b \in ℝ), if a^{2} + b^{2} =$

(a) 1
(b) −1
(c) 2
(d) −2

Answer:

$a - i b = \frac{3 - 4 i x}{3 + 4 i x} = \frac{3 - 4 i x}{3 + 4 i x} \times \frac{3 - 4 i x}{3 - 4 i x} = \frac{9 + 16 x^{2} i^{2} - 24 x i}{9 - 16 x^{2} i^{2}} = \frac{(9 - 16 x^{2}) - i (24 x)}{9 + 16 x^{2}} \Rightarrow {|a - i b|}^{2} = {|\frac{(9 - 16 x^{2}) - i (24 x)}{9 + 16 x^{2}}|}^{2} \Rightarrow a^{2} + b^{2} = \frac{{(9 - 16 x^{2})}^{2} + {(24 x)}^{2}}{{(9 + 16 x^{2})}^{2}} = \frac{81 + 256 x^{4} - 288 x^{2} + 576 x^{2}}{{(9 + 16 x^{2})}^{2}} = \frac{81 + 256 x^{4} + 288 x^{2}}{{(9 + 16 x^{2})}^{2}} = \frac{{(9 + 16 x^{2})}^{2}}{{(9 + 16 x^{2})}^{2}} = 1$

Hence, the correct option is (a).

Page No 13.65:

Question 40:

The complex number z which satisfies the condition $|\frac{i + z}{i - z}| = 1$ lies on

(a) circle x² + y² = 1
(b) the x−axis
(c) the y−axis
(d) the line x + y = 1

Answer:

$|\frac{i + z}{i - z}| = 1 \Rightarrow {|\frac{i + z}{i - z}|}^{2} = 1^{2} \Rightarrow (\frac{i + z}{i - z}) \bar{(\frac{i + z}{i - z})} = 1 \Rightarrow (\frac{i + z}{i - z}) (\frac{- i + \bar{z}}{- i - \bar{z}}) = 1 \Rightarrow (\frac{- i^{2} - z i + \bar{z} i + z \bar{z}}{- i^{2} + z i - \bar{z} i + z \bar{z}}) = 1 \Rightarrow - i^{2} - z i + \bar{z} i + z \bar{z} = - i^{2} + z i - \bar{z} i + z \bar{z} \Rightarrow - z i + \bar{z} i = z i - \bar{z} i \Rightarrow \bar{z} i + \bar{z} i = z i + z i \Rightarrow 2 \bar{z} i = 2 z i \Rightarrow \bar{z} = z \Rightarrow z is purely real$

Hence, the correct option is (b).

Page No 13.65:

Question 41:

If z is a complex number, then

(a) ${|z|}^{2} > {|z|}^{2}$
(b) ${|z|}^{2} = {|z|}^{2}$
(c) ${|z|}^{2} < {|z|}^{2}$
(d) ${|z|}^{2} \geq {|z|}^{2}$

Answer:

It is obvious that, for any complex number z,
${|z|}^{2} = {|z|}^{2}$

Hence, the correct option is (b).

Page No 13.65:

Question 42:

Which of the following is correct for any two complex numbers z₁ and z₂?

(a) $|z_{1} z_{2}| = |z_{1}| |z_{2}|$
(b) $\arg (z_{1} z_{2}) = \arg (z_{1}) \arg (z_{2})$
(c) $|z_{1} + z_{2}| = |z_{1}| + |z_{2}|$
(d) $|z_{1} + z_{2}| \geq |z_{1}| + |z_{2}|$

Answer:

Since, it is known that
$|z_{1} z_{2}| = |z_{1}| |z_{2}|$ ,
$\arg (z_{1} z_{2}) = \arg (z_{1}) + \arg (z_{2})$ and
$|z_{1} + z_{2}| \leq |z_{1}| + |z_{2}|$

Hence, the correct option is (a).

Page No 13.65:

Question 43:

If the complex number $z = x + i y$ satisfies the condition $|z + 1| = 1$ , then z lies on

(a) x−axis
(b) circle with centre (−1, 0) and radius 1
(c) y−axis
(d) none of these

Answer:

$|z + 1| = 1 \Rightarrow {|z + 1|}^{2} = 1^{2} \Rightarrow (z + 1) \bar{(z + 1)} = 1 \Rightarrow (z + 1) (\bar{z} + 1) = 1 \Rightarrow z \bar{z} + z + \bar{z} + 1 = 1 \Rightarrow z \bar{z} + z + \bar{z} = 0 Since, z = x + i y ∴ z \bar{z} + z + \bar{z} = 0 \Rightarrow (x + i y) (x - i y) + x + i y + x - i y = 0 \Rightarrow x^{2} + y^{2} + 2 x = 0 \Rightarrow {(x + 1)}^{2} + {(y - 0)}^{2} = 1^{2} which is the equation of a circle with centre (- 1, 0) and radius 1$

Hence, the correct option is (b).

Page No 13.65:

Question 44:

sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for

(a) x = nπ

(b) $x = (n + \frac{1}{2}) \frac{π}{2}$

(c) x = 0

(d) No value of x

Answer:

Given sin x + i cos 2x and cos x – i sin 2x are conjugate to each other
i.e sin x + i cos 2x = cos x – i sin 2x
i.e sin x – i cos 2x = cos x – i sin 2x
on comparing real and imaginary part,
sin x = cos x and cos 2x = sin 2x
i.e. sin x = cos x and 2cos²x – 1 = 2 sin x cos x
i.e 2cos²x – 1 = 2 cos x cos x (∴ sin x = cos x)
i.e 2cos²x – 1 = 2cos²x
i.e – 1 = 0
which is a false statement.
Hence no value of x exist
Therefore, the correct answer is option D.

Page No 13.66:

Question 45:

The real value of α for which the expression $\frac{1 - i \sin α}{1 + 2 i \sin α}$ is purely real, is

(a) $(n + 1) \frac{π}{2}$

(b) $(2 n + 1) \frac{π}{2}$

(c) nπ

(d) none of these where n ∈ N.

Answer:

Given $\frac{1 - i \sin α}{1 + 2 i \sin α}$ is purely real
$i . e \frac{1 - i \sin α}{1 + 2 i \sin α} \times (\frac{1 - 2 i \sin α}{1 - 2 i \sin α}) = \frac{1 - i \sin α - 2 i \sin α + 2 i^{2} \sin^{2} α}{1 - 4 i^{2} \sin^{2} α} = \frac{1 - 3 i \sin α - 2 \sin^{2} α}{1 + 4 \sin^{2} α} = \frac{1 - 2 \sin^{2} α}{1 + 4 \sin^{2} α} + i (\frac{- 3 \sin α}{1 + 4 \sin^{2} α})$

Which is given to purely real

$\Rightarrow \frac{- 3 \sin α}{1 + 4 \sin^{2} α} = 0 \Rightarrow - 3 \sin α = 0 i . e \sin α = 0 i . e α = n π$
Hence, the correct answer is option C.

Page No 13.66:

Question 46:

The value of $(z + 3) (\bar{z} + 3)$ is equivalent to

(a) |z + 3|²

(b) |z – 3|

(c) z² + 3

(d) none of these

Answer:

Since |z + 3|²

^{$= (z + 3) (z + 3) = (z + 3) (z + 3) (∵ \bar{3} = 3 real) = (z + 3) (z + 3)$}

Hence, the correct answer is option A.

Page No 13.66:

Question 47:

If ${(\frac{1 + i}{1 - i})}^{n} = 1,$ then n =
(a) 2m + 1
(b) 4m
(c) 2m
(d) 4m + 1 where m ∈ N

Answer:

Given :- ${(\frac{1 + i}{1 - i})}^{n} = 1$
$\begin{array}{rcl} Since, \frac{1 + i}{1 - i} & = & \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} \\ = & \frac{{(1 + i)}^{2}}{1 - i^{2}} \\ = & \frac{1 + i^{2} + 2 i}{1 + 1} \end{array}$

$i . e \frac{1 + i}{1 - i} = \frac{1 - 1 + 2 i}{2} = \frac{2 i}{2} = 2 i \Rightarrow {(\frac{1 + i}{1 - i})}^{n} = 1 \Rightarrow {(i)}^{n} = 1 \Rightarrow n = 4 m where m \in N$

Hence, the correct answer is option B.

Page No 13.66:

Question 48:

The vector represented by the complex number 2 – i is rotated about the origin through an angle $\frac{π}{2}$ in the clockwise direction, the new position of point is

(a) 1 + 2i

(b) –1 –2i

(c) 2 + i

(d) –1 + 2i

Answer:

Given 2 – i is rotated by $\frac{π}{2}$ angle in the clockwise direction about the origin
$i . e θ = - \frac{π}{2}$
Let Z' denote the new position and Z denote the previous p
$H e n c e Z' = Z e^{i θ} = Z e^{- \frac{i π}{2}} i . e Z' = (2 - i) [\cos (\frac{- π}{2}) + 1 \sin (\frac{- π}{2})] (∵ e^{i θ} = \cos θ + i \sin θ) i . e Z' = (2 - i) [0 + i (- 1)] = (2 - i) (- i) = - 2 i + i^{2} i . e Z' = - 1 - 2 i$

Hence, the correct answer is option B.

Page No 13.66:

Question 49:

The real value of θ for which the expression $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is a real number, is

(a) $n π + \frac{π}{4}$

(b) $n π + {(- 1)}^{n} \frac{π}{4}$

(c) $2 n π \pm \frac{π}{2}$

(d) none of theses

Answer:

Given $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is a real number
$i . e \frac{1 + i \cos θ}{1 - 2 i \cos θ} \times \frac{1 + 2 i \cos θ}{1 + 2 i \cos θ} = \frac{(1 + i + \cos θ) (1 + 2 i \cos θ)}{1 - 4 i^{2} \cos^{2} θ} = \frac{1 + 2 i \cos θ + i \cos θ + 2 i^{2} \cos^{2} θ}{1 + 4 \cos^{2} θ} = \frac{1 - 2 \cos^{2} θ + 3 i \cos θ}{1 + 4 \cos^{2} θ} = \frac{1 - 2 \cos^{2} θ}{1 + 4 \cos^{2} θ} + i (\frac{3 \cos θ}{1 + 4 \cos^{2} θ})$

Since $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is a real number

$\Rightarrow \frac{3 \cos θ}{1 + 4 \cos^{2} θ} = 0 i . e \cos θ = 0 i . e θ = \frac{(2 n \pm 1) π}{2} = n π \pm \frac{π}{2}$

Hence, the correct answer is option C.

Page No 13.66:

Question 50:

|z₁ + z₂| = |z₁| + |z₂| is possible if

(a) $z_{2} = {\bar{z}}_{1}$

(b) $z_{2} = \frac{1}{z_{1}}$

(c) arg (z₁) = arg (z₂)

(d) |z₁| = |z₂|

Answer:

Since given |z₁ + z₂| = |z₁| + |z₂|
i.e |z₁ + z₂|² = (|z₁| + |z₂|)²
$i . e {|z_{1} + z_{2}|}^{2} = {(|z_{1}| + |z_{2}|)}^{2} Since, |z_{1}| = r_{1}, |z_{2}| = r_{2} where z_{1} = r_{1} {e^{i θ}}_{1} z_{2} = r_{2} {e^{i θ}}_{2} {|z_{1}|}^{2} + {|z_{2}|}^{2} + 2 |z_{1}| |z_{2}| \cos (θ_{1} - θ_{2}) = r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} i . e r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \cos (θ_{1} - θ_{2}) = r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} i . e 2 r_{1} r_{2} \cos (θ_{1} - θ_{2}) = 2 r_{2} i . e \cos (θ_{1} - θ_{2}) = 1 i . e θ_{1} - θ_{2} = 0 i . e c h g z_{1} = c h g z_{2}$
Hence, the correct answer is option C.

Page No 13.66:

Question 51:

The equation |z + 1 – i| = |z – 1 + i| represents a
(a) straight line
(b) circle
(c) parabola
(d) hyperbola

Answer:

|z + 1 – i| = |z – 1 + i|
Let z = x + iy ; where x denote real part and y denote imaginary part of z
$straight i. straight e space open vertical bar x plus i y plus 1 minus i close vertical bar equals open vertical bar x plus i y minus 1 plus i close vertical bar straight i. straight e space open vertical bar open parentheses x plus 1 close parentheses plus i open parentheses y minus 1 close parentheses close vertical bar equals open vertical bar open parentheses x minus 1 close parentheses plus i open parentheses 1 plus y close parentheses close vertical bar Since space open vertical bar z close vertical bar equals square root of x squared plus y squared end root straight i. straight e space open vertical bar open parentheses x plus 1 close parentheses plus i open parentheses y minus 1 close parentheses close vertical bar squared equals open vertical bar open parentheses x minus 1 close parentheses plus i open parentheses y plus 1 close parentheses close vertical bar squared straight i. straight e space open parentheses x plus 1 close parentheses squared plus open parentheses y minus 1 close parentheses squared equals open parentheses x minus 1 close parentheses squared plus open parentheses y plus 1 close parentheses squared straight i. straight e space x squared plus 2 x plus 1 plus y squared minus 2 y plus 1 space equals x squared minus 2 x plus 1 plus y squared plus 2 y plus 1 straight i. straight e space 4 x equals 4 y straight i. straight e space x equals y semicolon space which space represents space straight a space straight space line$

Hence, the correct answer is option A.

Page No 13.66:

Question 52:

The area of the triangle on the complex plane formed by the complex numbers z, –iz and z + iz is

(a) |z|²

(b) $| \bar{z} |^{2}$

(c) $\frac{1}{2} {|z|}^{2}$

(d) none of these

Answer:

For any complex number z, –iz represents complex number obtained by rotating z clockwise by $\frac{π}{2}$ angle.
Hence, z, –iz and z + iz represents a right angled triangle with sides z, –iz and hypotenus z + iz
∴ Area of triangle formed is
$= \frac{1}{2} |z \times (- i z)| = \frac{1}{2} |i| {|z|}^{2} = \frac{1}{2} {|z|}^{2}$
Hence, the correct answer is option C.

Page No 13.67:

Question 1:

The principal value of the argument of the complex number 1 –i is ____________.

Answer:

Since z = 1 – i = 1 + i(–1)
$here θ = \tan^{- 1} |\frac{- 1}{1}| θ = \tan^{- 1} 1 = \frac{π}{y}$
Since z lies in 4^th quadrant.
∴argument is given by $\frac{- π}{y}$

Page No 13.67:

Question 2:

The polar form of (i²⁵)³ is ____________.

Answer:

(i²⁵)³
= (i²⁴. i)³
Since i²⁴ = 1
= (i)³ = i². i = – i
i.e (i²⁵)³ = – i
Here modulus $r = \sqrt{0^{2} + 1^{2}} = 1$ and argument $θ = \tan^{- 1} |\frac{- 1}{0}| = \infty = - \frac{π}{2}$
$\Rightarrow - i has polar form = \cos (\frac{- π}{2}) + i \sin (\frac{- π}{2}) Since \cos (- θ) = \cos θ and \sin (- θ) = - \sin θ$
Polar form of –i is
$\cos θ - i \sin θ where θ = \frac{π}{2} i . e \cos \frac{π}{2} - i \sin \frac{π}{2}$

Page No 13.67:

Question 3:

The value of $\sqrt{- 25} \times \sqrt{- 9}$ is ____________.

Answer:

$\sqrt{- 25} \times \sqrt{- 9} = 5 i \times 3 i (∵ \sqrt{- 1} = i) = 15 i^{2} = - 15 ∴ value of \sqrt{- 25} \times \sqrt{- 9} is - 15$

Page No 13.67:

Question 4:

The complex number $\frac{{(1 - i)}^{3}}{1 - i^{3}}$ in polar form is ____________.

Answer:

$\frac{{(1 - i)}^{3}}{1 - i^{3}} = \frac{{(1 - i)}^{3}}{1 + i} (∵ i^{2} = - 1) = \frac{{(1 - i)}^{3}}{1 + i} \times \frac{1 - i}{1 - i} = \frac{{(1 - i)}^{4}}{1 - i^{2}} = \frac{1}{2} \{{(1 - i)}^{2} {(1 - i)}^{2}\} = \frac{1}{2} \{(1 + i^{2} - 2 i) (1 + i^{2} - 2 i)\} = \frac{1}{2} \{(1 - 1 - 2 i) (1 - 1 - 2 i)\} = \frac{1}{2} \{+ 2 i \times 2 i\} = \frac{1}{2} \times 4 i^{2} = - 2 = - 2 (\cos 0 + i \sin 0) ∵ \cos 0 = 1 and \sin 0 = 0 Hence \frac{{(1 - i)}^{3}}{1 - i^{3}} = 2 (cosπ + i sinπ)$

Page No 13.67:

Question 5:

The sum of the series i + i² + i³ +_____ upto 1000 terms is ____________.

Answer:

i + i² + i³ +_____ 1000 terms
i.e i + i² + i³ + i⁴_____ + i^1000

= i + i² – i + 1 + i⁵ _____ + i¹⁰⁰⁰
= i – 1 – i + 1 + i⁵ + _______ + i¹⁰⁰⁰
= 0 + i⁵ + i⁶ + ________ + i¹⁰⁰
Similarly, sum of next form terms is also zero. (âˆµ 1000 = 4(250)
i.e multiple of 4
Hence, i + i² + i³ +_____ i¹⁰⁰⁰ = 0

Page No 13.67:

Question 6:

The multiplicative inverse of (1 + i) is ____________.

Answer:

For z = 1 + i
Let us suppose multiplicative inverse of 1 + i is a + ib
then (1 + i) (a + ib) = 1
i.e a + ib + ai + i²b = 1
i.e a + ib + ia – b = 1
i.e a – b + i(a + b) = 1 + i0
On comparing, real and imaginary part, we get
a – b = 1 and a + b = 0
i.e a – b = 1 and a = – b
i.e a + a = 1
$i . e a = \frac{1}{2} b = - \frac{1}{2}$
i.e multiplicative inverse of 1 + i is $\frac{1}{2} - \frac{i}{2}$

Page No 13.67:

Question 7:

If |z| = 4 and $\arg (z) = \frac{5 π}{6},$ then z = ____________.

Answer:

If |z| = 4 and $\arg z = \frac{5 π}{6} = \frac{π - π}{6}$
$∴ z = |z| [\cos (\frac{5 π}{6}) + i \sin (\frac{5 π}{6})] = 4 [\cos (π - \frac{π}{6}) + i \sin (π - \frac{π}{6})] [Since \cos (π - θ) = - \cos θ and \sin (π - θ) = \sin θ] ∴ z = 4 [- \cos \frac{π}{6} + i \sin \frac{π}{6}] Since \cos \frac{π}{6} = \frac{\sqrt{3}}{2} \sin \frac{π}{6} = \frac{1}{2} \Rightarrow z = 4 [- \frac{\sqrt{3}}{2} + \frac{i}{2}] = 2 (- \sqrt{3} + i) Hence, z = - 2 \sqrt{3} + 2 i$

Page No 13.67:

Question 8:

If z₁ and z₂ are two complex numbers such that z₁ + z₂ is a real number, then z₂ = ____________.

Answer:

Given for two complex numbers, z₁ and z₂, we have z₁+ z₂ is real number
$Let z_{1} = x_{1} + i y_{1} z_{2} = x_{2} + i y_{2} \Rightarrow z_{1} + z_{2} = x_{1} + i y_{1} + x_{2} + i y_{2} i . e z_{1} + z_{2} = (x_{1} + x_{2}) + i (y_{1} + y_{2}) Since z_{1} + z_{2} is real \Rightarrow y_{1} + y_{2} = 0 i . e y_{1} = - y_{2} i . e z_{2} = x_{2} + i y_{2} = x_{2} - i y_{1}$

Page No 13.67:

Question 9:

For any non-zero complex number z, arg (z) + arg $(\bar{z})$ = ____________.

Answer:

For complex number z
Say z = x + iy = re^iθ where r = modulus of z, θ = argument of $\bar{z} = x - i y$
$\Rightarrow \bar{z} = r e^{- i θ}$
Let us arg z = θ
Since arg $\bar{z}$ = – arg z
⇒ arg z + arg $\bar{z}$ = θ + (–θ)
i.e arg z + arg $(\bar{z})$ = 0

Page No 13.67:

Question 10:

If |z + 4| ≤ 3, then the greatest and least values of |z + 1| are _______ and ____________.

Answer:

Given |z + 4| ≤ 3
here |z + 1| = |z + 1 + 3 – 3| = |z + 4 + (–3)|
Since |a + b| ≤ |a| + |b| ≤ |z + 4| + |–3| = |z + 4| + 3
≤ 3 + 3 (given)
hence maximum value of |z + 1| is 6
|z + 1| = |z + 4 –3|
Since |a – b| ≥ ||a| – |b|| ≥ –|a| + |b|
⇒ |z + 1| ≥ – |z + 4| + 3
Since |z + 4| ≤ 3
⇒ –|z + 4| ≥ –3
i.e |z + 1| ≥ –|z + 4| + 3 ≥ –3 + 3 = 0
Hence, minimum value of |z + 1| is 0.

Page No 13.67:

Question 11:

The modulus and argument of $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$ are _______ and _______ respectively.

Answer:

Since complex number is $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$
$Modulus is \sqrt{\sin^{2} \frac{π}{5} + {(1 - \cos \frac{π}{5})}^{2}} = \sqrt{\sin^{2} \frac{π}{5} + 1 + \cos^{2} \frac{π}{5} - 2 \cos \frac{π}{5}} = \sqrt{- 2 \cos \frac{π}{5} + 1 + 1} (∵ \cos^{2} \frac{π}{5} + \sin^{2} \frac{π}{5} = 1) = \sqrt{2 - 2 \cos \frac{π}{5}} = \sqrt{2 (1 - \cos \frac{π}{5})} = \sqrt{2 \times 2 \sin^{2} \frac{π}{10}} (∵ 1 - \cos 2 θ = 2 \sin^{2} θ) i . e modulus = 2 \sin \frac{π}{10}$
and argument is $\tan^{- 1} (\frac{1 - \cos \frac{π}{5}}{\sin \frac{π}{5}})$
$= \tan^{- 1} (\frac{2 \sin^{2} \frac{π}{10}}{2 \sin \frac{π}{10} \cos \frac{π}{10}}) [∵ \sin 2 θ = 2 \sin θ \cos θ 1 - \cos 2 θ = 2 \cos^{2} θ] = \tan^{1} (\frac{2 \sin \frac{π}{10}}{2 \cos \frac{π}{10}}) = \tan^{1} (\tan \frac{π}{10}) = \frac{π}{10}$

∴ argument is given by $\frac{π}{10}$
∴ modulus of $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5}) is 2 \sin \frac{π}{10}$ and argument is $\frac{π}{10}$ .

Page No 13.67:

Question 12:

If $|\frac{z - 2}{z + 2}| = \frac{π}{6},$ then the locus of z is ____________.

Answer:

Given $|\frac{z - 2}{z + 2}| = \frac{π}{6}$
$i . e |z - 2| = \frac{π}{6} |z + 2| i . e {|z - 2|}^{2} = \frac{π^{2}}{6^{2}} {|z + 2|}^{2} for z = x + i y i . e {|x + i y - 2|}^{2} = \frac{π^{2}}{36} {|x + i y + 2|}^{2} i . e {|(x - 2) + i y|}^{2} = \frac{π^{2}}{36} {|(x + 2) + i y|}^{2} i . e {(x - 2)}^{2} + y^{2} = \frac{π^{2}}{36} [{(x + 2)}^{2} + y^{2}] i . e x^{2} + y^{2} + 4 - 4 x = \frac{π^{2}}{36} [x^{2} + y^{2} + 4 x + 4] i . e (1 - \frac{π^{2}}{36}) x^{2} + (1 - \frac{π^{2}}{36}) y^{2} + 4 (1 - \frac{π^{2}}{36}) - 4 x (1 + \frac{π^{2}}{36}) = 0$
which defines locus of a circle.

Page No 13.67:

Question 13:

If |z + 2i| = |z – 2i|, then the locus of z is ____________.

Answer:

Given |z + 2i| = |z – 2i|
for z = |x + iy|
|x + iy + 2i| = |x + iy – 2i|
Squaring both sides, |x + i(y + 2)|² = |x + i (y – 2)|²
i.e x² + (y + 2)|² = x² + (y – 2)|²
i.e x² + y² + 4 + 4y = x²+ y² + 4 – 4y
i.e 8y = 0
i.e y = 0
∴ locus is perpendicular bisector of the segment joining (0, –2) and (0, 2)

Page No 13.67:

Question 14:

If |z + 2| = |z – 2|, then the locus of z is ____________.

Answer:

|z + 2| = |z – 2| for z = x + iy
i.e |x + iy + 2| = |x + iy – 2|
i.e |(x + 2) + iy| = |(x – 2) + iy|
Square both sides,
|(x + 2)| + iy|² = |(x – 2)| + iy|²
i.e (x + 2)² + y² = (x – 2)² + y²
i.e x² + 4 + 4x + y² = x² + 4 – 4x + y²
i.e fx = 0
i.e x = 0
Hence, locus is perpendicular bisector of the segment joining (–2, 0) and (2, 0).

Page No 13.67:

Question 15:

If z = –1 + $\sqrt{- 3}$ , then arg (z) = ____________.

Answer:

$z = - 1 + i \sqrt{- 3} x = - 1 and y = \sqrt{3} θ = \tan^{- 1} \frac{\sqrt{3}}{- 1} = \tan^{- 1} |- \sqrt{3}| θ = \frac{π}{3}$ ,
Since z lies in IV quadrant.
⇒ argument of z is π $- \frac{π}{3} = \frac{2 π}{3}$ .

Page No 13.67:

Question 16:

If x < 0 is a real number, then arg (x) = ____________.

Answer:

If x < 0
i.e z = x + i 0 and x is negative
$\Rightarrow θ = \tan^{- 1} |\frac{0}{x}| = \tan^{- 1} 0 = 0$
Hence, z lies in II quadrant chg z = π.

Page No 13.67:

Question 17:

The real value of 'a' for which 3i³ – 2ai² + (1 – a) i + 5 is real is ____________.

Answer:

3i³ – 2ai² + (1 – a) i + 5
i.e 3i² i – 2a(–1) + (1 – a) i + 5
i.e 3i + 2a + (1 – a) i + 5
i.e 2a + 5 + i(1 – a – 3)
i.e 2a + 5 + i (–a – 2)
Since the above expression is given to be real
⇒ –a – 2 = 0
⇒ a = – 2

Page No 13.67:

Question 18:

If |z| = 2 and arg (z) = $\frac{π}{4}$ , then z = ____________.

Answer:

for |z| = 2 = r arg z = $\frac{π}{4}$

z = r (cos(arg z) + i sin (arg z))

$i . e z = 2 (\cos \frac{π}{4} + i \sin \frac{π}{4}) = 2 (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) z = \sqrt{2} + i \sqrt{2} hence, z = \sqrt{2} (1 + i)$

Page No 13.67:

Question 19:

The value of ${(- \sqrt{- 1})}^{4 n - 3},$ where n ∈ N, is ____________.

Answer:

${(- \sqrt{- 1})}^{4 n - 3} = {(- i)}^{4 n - 3} = {(- i)}^{4 n} {(- i)}^{- 3} = 1 {(- i)}^{- 3} = {(\frac{1}{- i})}^{3} = - \frac{1}{i^{3}} = \frac{- 1}{i^{2} . i} = \frac{- 1}{- 1 (i)} = + \frac{1}{i} \times \frac{i}{i} = \frac{i}{i^{2}} Hence, {(- \sqrt{- 1})}^{4 n - 3} = - i$

Page No 13.67:

Question 20:

The locus of z satisfying $\arg (z) = \frac{π}{3}$ is ____________.

Answer:

Given $\arg (z) = \frac{π}{3}$ and for z = x + iy
$\frac{π}{3} = \tan^{- 1} (\frac{y}{x}) i . e \tan \frac{π}{3} = \frac{y}{x} i . e \sqrt{3} = \frac{y}{x} i . e y = \sqrt{3} x in I quadrant except origin$

Page No 13.67:

Question 21:

The conjugate of the complex number $\frac{1 - i}{1 + i}$ is ____________.

Answer:

$\frac{(1 - i)}{(1 + i)} i . e conjugate of \frac{1 - i}{1 + i} = \frac{(1 - i)}{(1 + i)} = \frac{1 + i}{1 - i} = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{{(1 + i)}^{2}}{1 - i^{2}} = \frac{{(1 + i)}^{2}}{1 + 1} = \frac{1}{2} (1 + i^{2} + 2 i) Conjugate of \frac{1 - i}{1 + i} = \frac{1}{2} (2 i) = i$

Page No 13.67:

Question 22:

If (2 + i) (2 + 2i) (2 + 3i) ...... (2 + ni) = x + iy, then 5.8.13...(4 + n²) = ____________.

Answer:

Given:- (2 + i) (2 + 2i) (2 + 3i) ...... (2 + ni) = x + iy
Taking modulus both sides, we get
|(2 + i) (2 + 2i) (2 + 3i) ...... (2 + ni) = |x + iy|
By squaring both sides, we get
${|2 + i|}^{2} {|2 + 2 i|}^{2} {|2 + 3 i|}^{2} . . . . . . {|2 + n i|}^{2} = {|x + i y|}^{2} i . e (4 + 1) (4 + 4) (4 + 9) . . . . . . (4 + n^{2}) = x^{2} + y^{2} i . e 5.8.13 . . . . . . (4 + n^{2}) = x^{2} + y^{2}$

Page No 13.67:

Question 23:

If the point representing a complex number lies in the third quadrant, then the point representing its conjugate lies in the ____________.

Answer:

If z lies in III quadrant.

i.e z = – x – iy ; x, y ≥ 0

$\Rightarrow \bar{z} = - \bar{x - i y} = - x + i y$

i.e $\bar{z}$ lies in II quadrant.

Page No 13.67:

Question 24:

The multiplication of a non-zero complex number by i rotates it through ____________ in the anti-clockwise direction.

Answer:

Let z = x + iy then iz = ix + i²y
i.e iz = ix – y
i.e iz = – y + ix
$\tan θ_{1} for z = x + i y, \tan θ_{1} = \frac{y}{x} \tan θ_{2} for i z = i x - y, \tan θ_{2} = \frac{x}{- y} Since \tan θ_{1} \times \tan θ_{2} = \frac{y}{x} \times \frac{x}{- y} = - 1$
i.e iz is rotated by 90°.

Page No 13.67:

Question 25:

The complex number cosθ + i sinθ __________ be zero for any θ.

Answer:

z = cosθ + i sinθ can never be zero for any θ because z = 0
⇒ cosθ = 0 and sin θ = 0
Since no such value of θ exists
⇒ cosθ + i sinθ can never be zero.

Page No 13.67:

Question 26:

The argument of the complex number $(- 1 + i \sqrt{3}) (1 + i) (\cos θ + i \sin θ)$ is ____________.

Answer:

$For z = (- 1 + i \sqrt{3}) (1 + i) (\cos θ + i \sin θ) \arg z = \arg (- 1 + i \sqrt{3}) + \arg (1 + i) + \arg (\cos θ + i \sin θ) \arg (- 1 + i \sqrt{3}) : - z_{1} = - 1 + i \sqrt{3} θ_{1} = \tan^{- 1} |(\frac{\sqrt{3}}{1})| θ = \frac{π}{3}$
Since z₁ lies in II quadrant
$\Rightarrow \arg z_{1} = π - \frac{π}{3} = \frac{2 π}{3} \arg (1 + i) : - z_{2} = 1 + i θ_{2} = \tan^{- 1} |\frac{1}{1}| = \tan^{- 1} 1 i . e θ_{2} = \frac{π}{4}$
Since z₂ lies in I quadrant
$\Rightarrow \arg z_{2} = \frac{π}{4} \arg z_{3} = \arg (\cos θ + i \sin θ) = θ \Rightarrow \arg z = \arg z_{1} + \arg z_{2} + \arg z_{3} \arg z = \frac{2 π}{3} + \frac{π}{4} + θ = |\frac{11 π}{12} + θ|$

Page No 13.67:

Question 27:

If a complex number coincides with its conjugate, then it lies on ____________.

Answer:

Let z = x + iy and $\bar{z} = \bar{x + i y}$
$\bar{z} = x - i y$
Since z = $\bar{z}$ (given)
⇒ x + iy = x – iy
⇒ iy = – iy
⇒ 2iy = 0
i.e y = 0
Then z lies an x-axis.

Page No 13.67:

Question 28:

The points representing the complex number z for which |z + 1| < |z – 1| lie on the left side of ____________.

Answer:

Given |z + 1| < |z – 1|

Let z = x + iy where x, y ∈R

$i . e |x + i y + 1| < |x + i y - 1| \Rightarrow |(x + 1) + i y| < |(x - 1) + i y|$

Squaring both sides, we get

{|(x + 1) + i y|}^{2} < {|(x - 1) + i y|}^{2} i . e {(x + 1)}^{2} + y^{2} < {(x - 1)}^{2} + y^{2} i . e x^{2} + 1 + 2 x + y^{2} < x^{2} + 1 - 2 x + y^{2} i . e 2 x < 0 i . e x < 0

Hence, |z + 1| < |z – 1| lies on left side of y-axis

Page No 13.68:

Question 29:

If three complex numbers z₁, z₂ and z₃ are in A.P., then points representing them lie on ____________.

Answer:

Since z₁, z₂ and z₃are in A.P
Hence 2z₂ = z₁+ z₃
$i . e z_{2} = \frac{z_{1} + z_{3}}{z_{2}}$
i.e z₂ is the mid-point of line joining z₁and z₃
⇒ z₁, z₂and z₃lie on a straight.

Page No 13.68:

Question 30:

The principal argument of i^–1097 is ____________.

Answer:

Let z = i^–1097
$z = \frac{1}{{(i)}^{1096} \cdot i} = \frac{1}{{(i^{4})}^{274}} \times \frac{1}{i} = \frac{1}{1} \times \frac{1}{i} (∵ i^{4} = 1) = \frac{1}{i} = \frac{1}{i} \times \frac{i}{i} z = - i = \cos (\frac{π}{2}) - i \sin \frac{π}{2}$
Hence, principle argument is $- \frac{π}{2}$ .

Page No 13.68:

Question 31:

The value of $\frac{i^{4 n + 1} - i^{4 n - 1}}{2}$ is ____________.

Answer:

$\begin{array}{rcl} Let z & = & \frac{i^{4 n + 1} - i^{4 n - 1}}{2} \\ = & \frac{i^{4 n} . i - i^{4 n} - i^{- 1}}{2} \\ = & \frac{i - i^{- 1}}{2} \\ = & \frac{i - \frac{1}{i}}{2} \\ = & \frac{i^{2} - 1}{2 i} \\ = & - \frac{1 - 1}{2 i} \\ = & - \frac{2}{2 i} \\ = & - \frac{1}{i} \times \frac{i}{i} \\ = & - \frac{i}{- 1} \end{array}$

$i . e \frac{i^{4 n + 1} - i^{4 n - 1}}{2} = i$

Page No 13.68:

Question 32:

If $z_{1} = \sqrt{3} + i \sqrt{3}$ and $z_{2} = \sqrt{3} + i,$ then the point representing $\frac{z_{1}}{z_{2}}$ lies in ____________.

Answer:

$Let z_{1} = \sqrt{3} (1 + i) and z_{2} = \sqrt{3} + i$

$\begin{array}{rcl} then \frac{z_{1}}{z_{2}} & = & \frac{\sqrt{3} (1 + i)}{\sqrt{3} + i} \times \frac{\sqrt{3} - i}{\sqrt{3} - i} \\ = & \frac{\sqrt{3} (1 + i) (\sqrt{3} - i)}{3 - i^{2}} \\ = & \frac{\sqrt{3} (\sqrt{3} - i + \sqrt{3} i - i^{2})}{3 + 1} \\ = & \frac{\sqrt{3}}{4} (\sqrt{3} + 1 + i (\sqrt{3} - 1)) \end{array}$

$Since \sqrt{3} + 1 and \sqrt{3} - 1 > 0 \Rightarrow \frac{z_{1}}{z_{2}} lies in I quadrant$

Page No 13.68:

Question 33:

If 0 < arg (z) < π, then arg (z) – arg (–z) = ____________.

Answer:

For 0 < arg z < π
Let z = r(cosθ, i sinθ)
i.e arg z = θ
Then –z = – r(cosθ + i sinθ)
= – r (+ cosθ + i (+ sinθ))

    = (–1) re^iθ
    = e^iπre^iθ
    = reⁱ⁽^{θ + π)}

i.e arg (–z) = θ + π
⇒ arg z – arg(–z) = θ – θ – π
= – π

Page No 13.68:

Question 34:

For any two complex numbers z₁, z₂ and any real numbers a, b, |az₁ – bz₂|² + |bz₁ + az₂|² = ____________.

Answer:

For complex z₁ and z₂and real numbers a and b
${|a z_{1} - b z_{2}|}^{2} = (a z_{1} - b z_{2}) (\bar{a z_{1} - b z_{2}}) = (a z_{1} - b z_{2}) (a {\bar{z}}_{1} - b {\bar{z}}_{2}) = a^{2} z_{1} {\bar{z}}_{1} - a b z_{1} {\bar{z}}_{2} - a b z_{2} {\bar{z}}_{1} + b^{2} z_{2} {\bar{z}}_{2} {|a z_{1} - b z_{2}|}^{2} = a^{2} {|z_{1}|}^{2} - a b z_{1} {\bar{z}}_{2} - a b {\bar{z}}_{1} z_{2} + b^{2} z_{2} {\bar{z}}_{2} . . . (1) and {|a z_{1} + a z_{2}|}^{2} = (b z_{1} + a z_{2}) (\bar{b z_{1} + a z_{2}}) = (b z_{1} + a z_{2}) (b {\bar{z}}_{1} + a {\bar{z}}_{2}) = b^{2} z_{1} {\bar{z}}_{1} + a b z_{1} {\bar{z}}_{2} + a b z_{2} {\bar{z}}_{1} + a^{2} z_{2} {\bar{z}}_{2} = b^{2} {|z_{1}|}^{2} + a b z_{1} {\bar{z}}_{2} + a b {\bar{z}}_{1} z_{2} + a^{2} {|z_{2}|}^{2} . . . (2) ∴ from (1) and (2), {|a z_{1} - b z_{2}|}^{2} + {(b z_{1} + a z_{2})}^{2} = (a^{2} + b^{2}) ({|z_{1}|}^{2}) + (a^{2} + b^{2}) {|z_{2}|}^{2} = (a^{2} + b^{2}) ({|z_{1}|}^{2} + {|z_{2}|}^{2})$

Page No 13.68:

Question 35:

Let z₁ and z₂ be two complex numbers such that |z₁ + z₂| = |z₁| + |z₂|, then arg (z₁) – arg (z₂) = ____________.

Answer:

Given for complex number z₁ and z_{2
$|z_{1} + z_{2}| = |z_{1}| + |z_{2}| i . e {|z_{1} + z_{2}|}^{2} = {(|z_{1}| + |z_{2}|)}^{2} i . e {|z_{1}|}^{2} + {|z_{2}|}^{2} + 2 |z_{1}| |z_{2}| \cos θ = {|z_{1}|}^{2} + {|z_{2}|}^{2} + 2 |z_{1}| |z_{2}| (∵ z_{1} . z_{2} = |z_{1}| |z_{2}| \cos θ where θ is angle between z_{1} and z_{2}) i . e 2 |z_{1}| |z_{2}| \cos θ = 2 |z_{1}| |z_{2}| i . e \cos θ = 1 i . e θ = 0$}
i.e angle between z₁ and z₂is 0
i.e arg (z₁) – arg z₂ = 0

Page No 13.68:

Question 36:

Let z₁ and z₂ be two complex numbers such that |z₁ + z₂| = |z₁ – z₂|, then arg (z₁) – arg (z₂) = ____________.

Answer:

Given |z₁ + z₂| = | z₁– z₂|
On squaring both sides, |z₁ + z₂|² = | z₁– z₂|² (1)
i.e for
$z subscript 1 equals r subscript 1 open parentheses cos theta subscript 1 plus i space sin theta subscript 1 close parentheses z subscript 2 equals r subscript 2 open parentheses cos theta subscript 2 plus i space sin theta subscript 2 close parentheses$
Using identities
$open vertical bar z subscript 1 plus z subscript 2 close vertical bar squared equals open vertical bar z subscript 1 close vertical bar squared plus open vertical bar z subscript 2 close vertical bar squared plus 2 open vertical bar z subscript 1 close vertical bar open vertical bar z subscript 2 close vertical bar space cos open parentheses theta subscript 1 minus theta subscript 2 close parentheses open vertical bar z subscript 1 minus z subscript 2 close vertical bar squared equals open vertical bar z subscript 1 close vertical bar squared plus open vertical bar z subscript 2 close vertical bar squared minus 2 open vertical bar z subscript 1 close vertical bar open vertical bar z subscript 2 close vertical bar space cos open parentheses theta subscript 1 minus theta subscript 2 close parentheses therefore space from space left parenthesis 1 right parenthesis comma straight i. straight e space open vertical bar z subscript 1 close vertical bar squared plus open vertical bar z subscript 2 close vertical bar squared plus 2 open vertical bar z subscript 1 close vertical bar open vertical bar z subscript 2 close vertical bar space cos open parentheses theta subscript 1 minus theta subscript 2 close parentheses space equals open vertical bar z subscript 1 close vertical bar squared plus open vertical bar z subscript 2 close vertical bar squared minus 2 open vertical bar z subscript 1 close vertical bar open vertical bar z subscript 2 close vertical bar space cos open parentheses theta subscript 1 minus theta subscript 2 close parentheses straight i. straight e space cos open parentheses theta subscript 1 minus theta subscript 2 close parentheses equals 0 rightwards double arrow theta subscript 1 minus theta subscript 2 equals straight pi over 2 straight i. straight e space arg space z subscript 1 minus arg space z subscript 2 equals straight pi over 2$

Page No 13.68:

Question 37:

If |z₁| = |z₂| and arg $(\frac{z_{1}}{z_{2}}) = π,$ then z₁ + z₂ = ____________.

Answer:

Let |z₁| = |z₂| = r
Let arg z₁ = θ₁ and arg z₂ = θ₂
$\Rightarrow z_{1} = r e^{i θ_{1}} and z_{2} = r e^{i θ_{2}} Since \arg (\frac{z_{1}}{z_{2}}) = π i . e \arg z_{1} - \arg z_{2} = π i . e θ_{1} - θ_{2} = π i . e θ_{1} = θ_{2} + π e^{i θ_{1}} = e^{i (θ_{2} + π)} = e^{i π} e^{i θ_{2}} = (cosπ + i sinπ) (\cos θ_{2} + i \sin θ_{2}) = - 1 (\cos θ_{2} + i \sin θ_{2}) e^{i θ_{1}} = - e^{i θ_{2}} \Rightarrow z_{1} + z_{2} = r e^{i θ_{1}} + r e^{i θ_{2}} Hence, z_{1} + z_{2} = r (e^{i θ_{1}} + e^{i θ_{2}}) = 0$

Page No 13.68:

Question 1:

Write the values of the square root of i.

Answer:

$Let the square root of i be x + i y . \Rightarrow \sqrt{i} = x + i y \Rightarrow i = x^{2} + y^{2} i^{2} + 2 i x y \Rightarrow i = x^{2} - y^{2} + 2 i x y (Squaring both the sides) Comparing both the sides : x^{2} - y^{2} = 0 . . . (i) and 2 x y = 1 . . . (ii) By equation (ii), we find that x and y are of the same sign . From equation (i), x = \pm y ∴ x y = \frac{1}{2}, x^{2} = \frac{1}{2} x = \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{2}} ∴ \sqrt{i} = \pm \frac{1}{\sqrt{2}} (1 + i)$

Page No 13.68:

Question 2:

Write the values of the square root of −i.

Answer:

$Let \sqrt{- i} = x + i y Squaring both the sides - i = x^{2} + y^{2} i^{2} + 2 i x y \Rightarrow 2 x y = - 1 . . . (i) and x^{2} - y^{2} = 0 . . . (ii) Equation (ii) shows that x and y are of opposite sign . From (ii), x = \pm y From (i), 2 (x) (- x) = \frac{- 1}{2} \Rightarrow x^{2} = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}} [Since x and y have opposite signs, y = - \frac{1}{\sqrt{2}} when x = \frac{1}{\sqrt{2}} and vice versa] ∴ \sqrt{- i} = \pm \frac{1}{\sqrt{2}} (1 - i)$

Page No 13.69:

Question 3:

If x + iy = $\sqrt{\frac{a + i b}{c + i d}}$ , then write the value of (x² + y²)².

Answer:

$x + i y = \sqrt{\frac{a + i b}{c + i d}} Taking modulus on both the sides, |x + i y| = |\sqrt{\frac{a + i b}{c + i d}}| \Rightarrow |x + i y| = \sqrt{\frac{|a + i b|}{|c + i d|}} \Rightarrow \sqrt{x^{2} + y^{2}} = \sqrt{\frac{\sqrt{a^{2} + b^{2}}}{\sqrt{c^{2} + d^{2}}}} [∵ |x + i y| = \sqrt{x^{2} + y^{2}}] Squaring both the sides, x^{2} + y^{2} = \sqrt{\frac{a^{2} + b^{2}}{c^{2} + d^{2}}} Squaring again, we get, {(x^{2} + y^{2})}^{2} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}}$

Page No 13.69:

Question 4:

If π < θ < 2π and z = 1 + cos θ + i sin θ, then write the value of $|z|$ .

Answer:

π < θ < 2 π \frac{π}{2} < \frac{θ}{2} < π (Dividing by 2) z = 1 + \cos θ + isin θ \Rightarrow |z| = \sqrt{{(1 + \cos θ)}^{2} + \sin^{2} θ} \Rightarrow |z| = \sqrt{1 + \cos^{2} θ + 2 \cos θ + \sin^{2} θ} \Rightarrow |z| = \sqrt{1 + 1 + 2 \cos θ} \Rightarrow |z| = \sqrt{2 (1 + \cos θ)} \Rightarrow |z| = \sqrt{2 \times 2 \cos^{2} \frac{θ}{2}} \Rightarrow |z| = 2 \sqrt{\cos^{2} \frac{θ}{2}} \Rightarrow |z| = - 2 \cos \frac{θ}{2} [Since \frac{π}{2} < \frac{θ}{2} < π, \cos \frac{θ}{2} is negative]

Page No 13.69:

Question 5:

If n is any positive integer, write the value of $\frac{i^{4 n + 1} - i^{4 n - 1}}{2}$ .

Answer:

$\frac{i^{4 n + 1} - i^{4 n - 1}}{2} = \frac{i - \frac{1}{i}}{2} (∵ i^{4 n} = 1, i^{- 1} = \frac{1}{i}) = \frac{\frac{i^{2} - 1}{i}}{2} = \frac{i^{2} - 1}{2 i} = \frac{- 1 - 1}{2 i} = \frac{- 2}{- 2 i} = \frac{- 1}{i} = \frac{- i}{i^{2}} (∵ i^{2} = - 1) = \frac{- i}{- 1} = i$

Page No 13.69:

Question 6:

Write the value of $\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}}$ .

Answer:

$\frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} = \frac{i^{4 \times 148} + i^{4 \times 147 + 2} + i^{4 \times 147} + i^{146 \times 4 + 2} + i^{4 \times 146}}{i^{4 \times 145 + 2} + i^{4 \times 145} + i^{4 \times 144 + 2} + i^{4 \times 144} + i^{4 \times 143 + 2}} = \frac{1 + i^{2} + 1 + i^{2} + 1}{i^{2} + 1 + i^{2} + 1 + i^{2}} [∵ i^{4} = 1] = \frac{1 - 1 + 1 - 1 + 1}{- 1 + 1 - 1 + 1 - 1} [∵ i^{2} = 1] = \frac{1}{- 1} = - 1$

Page No 13.69:

Question 7:

Write 1 − i in polar form.

Answer:

$z = 1 - i r = |z| = \sqrt{1 + 1} = \sqrt{2} Let \tan α = |\frac{Im (z)}{Re (z)}| ∴ \tan α = |\frac{- 1}{1}| = \frac{π}{4} \Rightarrow α = \frac{π}{4} Since point (1, - 1) lies in the fourth quadrant, the argument of z is given by θ = - α = - \frac{π}{4} Polar form = r (\cos θ + i \sin θ) = \sqrt{2} \{\cos (- \frac{π}{4}) + i \sin (- \frac{π}{4})\} = \sqrt{2} (\cos \frac{π}{4} - i \sin \frac{π}{4})$

Page No 13.69:

Question 8:

Write −1 + i $\sqrt{3}$ in polar form

Answer:

$Let z = - 1 + \sqrt{3} i . T h e n, r = |z| = \sqrt{{[- 1]}^{2} + {[\sqrt{3}]}^{2}} = 2 Let \tan α = |\frac{Im (z)}{Re (z)}| = \sqrt{3} \Rightarrow α = \frac{π}{3} Since the point representing z lies in the second quadrant . Therefore, the argument of z is given by θ = π - α = π - \frac{π}{3} = \frac{2 π}{3} So, the polar form is r (\cos θ + i \sin θ) ∴ z = 2 (\cos \frac{2 π}{3} + i \sin \frac{2 π}{3})$

Page No 13.69:

Question 9:

Write the argument of −i.

Answer:

$Let z = - i Then, Re (z) = 0, Im (z) = - 1 Since, the point (0, - 1) representing z = 0 - i lies on negative direction of imaginary axis . Therefore, \arg (z) = \frac{- π}{2} or \frac{3 π}{2}$

Page No 13.69:

Question 10:

Write the least positive integral value of n for which ${(\frac{1 + i}{1 - i})}^{n}$ is real.

Answer:

${(\frac{1 + i}{1 - i})}^{n} = {(\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i})}^{n} = {(\frac{1 + i^{2} + 2 i}{1 - i^{2}})}^{n} = {(\frac{1 - 1 + 2 i}{1 + 1})}^{n} \Rightarrow {(\frac{2 i}{2})}^{n} = i^{n} For i^{n} to be real, the smallest positive value of n will be 2 . As, i^{2} = - 1, which is real .$

Page No 13.69:

Question 11:

Find the principal argument of ${(1 + i \sqrt{3})}^{2}$ .

Answer:

$z = {(1 + i \sqrt{3})}^{2} = 1 + 3 i^{2} + 2 \sqrt{3} i = 1 - 3 + 2 \sqrt{3} i = - 2 + 2 \sqrt{3} i$

$Let β be an acute angle given by \tan β = \frac{|Im (z)|}{|Re (z)|} . Then, \tan β = \frac{|2 \sqrt{3}|}{|2|} = |\sqrt{3}| \Rightarrow \tan β = |\tan \frac{π}{3}| \Rightarrow β = \frac{π}{3} Clearly, z lies in the second quadrant . Therefore, \arg (z) = π - \frac{π}{3} = \frac{2 π}{3} . Hence, the principal argument of z is \frac{2 π}{3} .$

Page No 13.69:

Question 12:

Find z, if $|z| = 4 and \arg (z) = \frac{5 π}{6} .$

Answer:

We know that,

$z = |z| \{\cos [\arg (z)] + i \sin [\arg (z)]\} \Rightarrow z = 4 (\cos \frac{5 π}{6} + i \sin \frac{5 π}{6}) = 4 (- \cos \frac{π}{6} + i \sin \frac{π}{6}) = 4 (- \frac{\sqrt{3}}{2} + \frac{1}{2} i) = - 2 \sqrt{3} + 2 i$

Thus, $z = - 2 \sqrt{3} + 2 i$ .

Page No 13.69:

Question 13:

If $|z - 5 i| = |z + 5 i|$ , then find the locus of z.

Answer:

$|z - 5 i| = |z + 5 i| \Rightarrow {|z - 5 i|}^{2} = {|z + 5 i|}^{2} \Rightarrow (z - 5 i) (\bar{z - 5 i}) = (z + 5 i) (\bar{z + 5 i}) [∵ z \bar{z} = {|z|}^{2}] \Rightarrow (z - 5 i) (\bar{z} + 5 i) = (z + 5 i) (\bar{z} - 5 i) \Rightarrow z \bar{z} + 5 z i - 5 \bar{z} i - 25 i^{2} = z \bar{z} - 5 z i + 5 \bar{z} i - 25 i^{2} \Rightarrow 5 z i + 5 z i = 5 \bar{z} i + 5 \bar{z} i \Rightarrow 10 z i = 10 \bar{z} i \Rightarrow z = \bar{z} \Rightarrow z is purely real$

Hence, the locus of z is real axis.

Page No 13.69:

Question 14:

If $\frac{{(a^{2} + 1)}^{2}}{2 a - i} = x + i y$ , find the value of $x^{2} + y^{2}$ .

Answer:

$\frac{{(a^{2} + 1)}^{2}}{2 a - i} = x + i y . . . . (1) \Rightarrow [\bar{\frac{{(a^{2} + 1)}^{2}}{2 a - i}}] = \bar{x + i y} \Rightarrow \frac{{(a^{2} + 1)}^{2}}{2 a + i} = x - i y . . . . (2) On multiplying (1) and (2), we get \frac{{(a^{2} + 1)}^{2}}{2 a - i} \times \frac{{(a^{2} + 1)}^{2}}{2 a + i} = (x + i y) (x - i y) \Rightarrow \frac{{(a^{2} + 1)}^{4}}{{(2 a)}^{2} - i^{2}} = x^{2} - i^{2} y^{2} \Rightarrow \frac{{(a^{2} + 1)}^{4}}{{(2 a)}^{2} + 1} = x^{2} + y^{2}$

Hence, $x^{2} + y^{2} = \frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1}$ .

Page No 13.69:

Question 15:

Write the value of $\sqrt{- 25} \times \sqrt{- 9}$ .

Answer:

$\sqrt{- 25} \times \sqrt{- 9} = 5 \sqrt{- 1} \times 3 \sqrt{- 1} = 5 i \times 3 i = 15 i^{2} = - 15$

Hence, $\sqrt{- 25} \times \sqrt{- 9} = - 15$ .

Page No 13.69:

Question 16:

Write the sum of the series $i + i^{2} + i^{3} + . . . .$ upto 1000 terms.

Answer:

We know that,
$i + i^{2} + i^{3} + i^{4} = i - 1 - i + 1 = 0$

$∴ i + i^{2} + i^{3} + . . . . + i^{1000} = (i + i^{2} + i^{3} + i^{4}) + (i^{5} + i^{6} + i^{7} + i^{8}) + . . . + (i^{997} + i^{998} + i^{999} + i^{1000}) = (i + i^{2} + i^{3} + i^{4}) + (i^{4} i + i^{4} i^{2} + i^{4} i^{3} + i^{4} i^{4}) + . . . + [{(i^{4})}^{249} i + {(i^{4})}^{249} i^{2} + {(i^{4})}^{249} i^{3} + {(i^{4})}^{249} i^{4}] = (i + i^{2} + i^{3} + i^{4}) + (i + i^{2} + i^{3} + i^{4}) + . . . + (i + i^{2} + i^{3} + i^{4}) = 0$

Thus, the sum of the series $i + i^{2} + i^{3} + . . . .$ upto 1000 terms is 0.

Page No 13.69:

Question 17:

Write the value of $\arg (z) + \arg (\bar{z})$ .

Answer:

Let z be a complex number with argument θ.
Then,
$z = r e^{i θ} \Rightarrow \bar{z} = \bar{r e^{i θ}} = r e^{- i θ}$
⇒ argument of $\bar{z}$ is −θ.

Thus, $\arg (z) + \arg (\bar{z}) = 0$ .

Page No 13.69:

Question 18:

If $|z + 4| \leq 3$ , then find the greatest and least values of $|z + 1|$ .

Answer:

$|z + 1| = |z + 4 - 3| \leq |z + 4| + |- 3| \leq 3 + 3 = 6 Also, |z + 1| \geq 0 Thus, 0 \leq |z + 1| \leq 6 .$

Hence, the greatest and least values of $|z + 1|$ is 6 and 0.

Page No 13.69:

Question 19:

For any two complex numbers z₁ and z₂ and any two real numbers a, b, find the value of ${|a z_{1} - b z_{2}|}^{2} + {|a z_{2} + b z_{1}|}^{2}$ .

Answer:

${|a z_{1} - b z_{2}|}^{2} + {|a z_{2} + b z_{1}|}^{2} = (a z_{1} - b z_{2}) (\bar{a z_{1} - b z_{2}}) + (a z_{2} + b z_{1}) (\bar{a z_{2} + b z_{1}}) = (a z_{1} - b z_{2}) (a \bar{z_{1}} - b \bar{z_{2}}) + (a z_{2} + b z_{1}) (a \bar{z_{2}} + b \bar{z_{1}}) = (a^{2} z_{1} \bar{z_{1}} - a b z_{1} \bar{z_{2}} - a b z_{2} \bar{z_{1}} + b^{2} z_{2} \bar{z_{2}}) + (a^{2} z_{2} \bar{z_{2}} + a b z_{1} \bar{z_{2}} + a b z_{2} \bar{z_{1}} + b^{2} z_{1} \bar{z_{1}}) = [(a^{2} + b^{2}) z_{1} \bar{z_{1}} + (a^{2} + b^{2}) z_{2} \bar{z_{2}}] = [(a^{2} + b^{2}) (z_{1} \bar{z_{1}} + z_{2} \bar{z_{2}})] = [(a^{2} + b^{2}) ({|z_{1}|}^{2} + {|z_{2}|}^{2})]$

Hence, ${|a z_{1} - b z_{2}|}^{2} + {|a z_{2} + b z_{1}|}^{2} = (a^{2} + b^{2}) ({|z_{1}|}^{2} + {|z_{2}|}^{2})$ .

Page No 13.69:

Question 20:

Write the conjugate of $\frac{2 - i}{{(1 - 2 i)}^{2}}$ .

Answer:

$\frac{2 - i}{{(1 - 2 i)}^{2}} = \frac{2 - i}{1 + 4 i^{2} - 4 i} = \frac{2 - i}{1 - 4 - 4 i} = \frac{2 - i}{- 3 - 4 i} = \frac{- 2 + i}{3 + 4 i} = \frac{i - 2}{3 + 4 i} \times \frac{3 - 4 i}{3 - 4 i} = \frac{3 i - 4 i^{2} - 6 + 8 i}{3^{2} - 4^{2} i^{2}} = \frac{11 i + 4 - 6}{9 + 16} = \frac{- 2}{25} + \frac{11}{25} i$

∴ Conjugate of $\frac{2 - i}{{(1 - 2 i)}^{2}} = (\bar{- \frac{2}{25} + \frac{11}{25} i}) = - \frac{2}{25} - \frac{11}{25} i$

Hence, Conjugate of $\frac{2 - i}{{(1 - 2 i)}^{2}}$ is $- \frac{2}{25} - \frac{11}{25} i$ .

Page No 13.69:

Question 21:

If n ∈ $ℕ$ , then find the value of $i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3}$ .

Answer:

$i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} = i^{n} + i^{n} . i + i^{n} . i^{2} + i^{n} . i^{3} = i^{n} + i^{n} . i + i^{n} . (- 1) + i^{n} . (- i) = i^{n} + i^{n} . i - i^{n} - i^{n} . i = 0$

Thus, the value of $i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3}$ is 0.

Page No 13.69:

Question 22:

Find the real value of a for which $3 i^{3} - 2 a i^{2} + (1 - a) i + 5$ is real.

Answer:

$3 i^{3} - 2 a i^{2} + (1 - a) i + 5 = - 3 i + 2 a + (1 - a) i + 5 = (2 a + 5) + i (1 - a - 3) = (2 a + 5) + i (- 2 - a)$

Since, $3 i^{3} - 2 a i^{2} + (1 - a) i + 5$ is real.

$∴ Im [3 i^{3} - 2 a i^{2} + (1 - a) i + 5] = 0 \Rightarrow - 2 - a = 0 \Rightarrow a = - 2$

Hence, the real value of a for which $3 i^{3} - 2 a i^{2} + (1 - a) i + 5$ is real is −2.

Page No 13.69:

Question 23:

If $|z| = 2 and \arg (z) = \frac{π}{4}$ , find z.

Answer:

We know that,
$z = |z| \{c o s [\arg (z)] + i \sin [\arg (z)]\} = 2 (\cos \frac{π}{4} + i \sin \frac{π}{4}) = 2 (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) = \sqrt{2} (1 + i)$

Hence, $z = \sqrt{2} (1 + i)$ .

Page No 13.69:

Question 24:

Write the argument of $(1 + i \sqrt{3}) (1 + i) (\cos θ + i \sin θ)$ .

Disclaimer: There is a misprinting in the question. It should be $(1 + i \sqrt{3})$ instead of $(1 + \sqrt{3})$ .

Answer:

Let the argument of $(1 + i \sqrt{3})$ be α. Then,
$\tan α = \frac{\sqrt{3}}{1} = \tan \frac{π}{3} \Rightarrow α = \frac{π}{3}$

Let the argument of $(1 + i)$ be β. Then,
$tanβ = \frac{1}{1} = \tan \frac{π}{4} \Rightarrow β = \frac{π}{4}$

Let the argument of $(cosθ + i sinθ)$ be γ. Then,
$tanγ = \frac{sinθ}{cosθ} = tanθ \Rightarrow γ = θ$

∴ The argument of $(1 + i \sqrt{3}) (1 + i) (cosθ + i sinθ) = α + β + γ = \frac{π}{3} + \frac{π}{4} + θ = \frac{7 π}{12} + θ$

Hence, the argument of $(1 + i \sqrt{3}) (1 + i) (cosθ + i sinθ) is \frac{7 π}{12} + θ$ .

View NCERT Solutions for all chapters of Class 15

Login or Create a free account