Solve this: If ak 3, 1≤k≤n, then all complex number z satisfying equation 1+a1z+a2z2+ - Maths - Complex Numbers and Quadratic Equations

Question

Solve this:
I f   a k < 3 ,   1 ≤ k ≤ n ,   t h e
n   a l l   c o m p l e x   n u m b e r   z
  s a t i s f y i n g   e q u a t i o n   1 + a 1 z
+ a 2 z 2 + + a n z n = 0 ( a )   l i e   o u t s i d e
  c i r c l e   z = 1 / 4        
                  ( b
)   l i e   i n s i d e   c i r c l e   z = 1 /
4 ( c )   l i e   o n   t h e   c i r c l e
  z = 1 / 4              
                ( d )
  l i e   i n   1 / 3 < z < 1 / 2

Shivam Mishra · Accepted Answer

Dear Student,
Please find below the solution to the asked query:

We have1+a1z+a2z2+ anzn=0⇒a1z+a2z2+
anzn=-1⇒∑r=1n arzr=1Using modulus inequality we get⇒∑r=1n arzr≤∑r=1narzr⇒1≤∑r=1narzrAs ar<3 ⇒1<∑r=1n 3zr Equality will be removed as we have put limiting maximumvalue of ar⇒1<∑r=1n 3zr⇒∑r=1nzr>13Hence option d is correct

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