prove the triangle inequality: IZ1 - z2I or equal to Iz1I+Iz2I - Maths - Complex Numbers and Quadratic Equations

Question

prove the triangle inequality: IZ1 - z2I <or equal to
Iz1I+Iz2I

Chetna Khanna · Accepted Answer

Triangle inequality states that: $| z_{1} - z_{2} | \geq | z_{1} | - | z_{2} |$ Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers Let z1 and z2 be represented by the prints P(x1, y1) and Q(x2, y2) in the arg and plane Let O be the origin Join OP and OQ Produce QO backwards to Q' such that OQ' = OQ Coordinates of Q' will be (â€“ x2, â€“ y2) So the complex number represented by Q' is â€“ z2 = â€“ x2 + i(â€“ y2) OPRQ' will form a parallelogram and its diagonals will intersects at point H H is the middle point of PQ', So coordinates of H are $(\frac{x_{1} - x_{2}}{2}, \frac{y_{1} - y_{2}}{2})$ Let (x, y) be the coordinates of R As, H is the mid point of OR âˆ´ Coordinates of H are $(\frac{x}{2}, \frac{y}{2})$

$∴ \frac{x}{2} = \frac{x_{1} - x_{2}}{2} \Rightarrow x = x_{1} - x_{2} and \frac{y}{2} = \frac{y_{1} - y_{2}}{2} \Rightarrow y = y_{1} - y_{2}$ So, coordinates of R are (x1 â€“ x2, y1 â€“ y2) Complex number represented by R = (x1 â€“ x2) + i (y1 â€“ y2) = (x1 + iy1) â€“ (x2 + iy2) = z1 = z2 $| z_{1} | = OP, | z_{2} | = OQ' and | z_{1} - z_{2} | = OR$ As the difference of any two sides of a triangle is less than the third side âˆ´ OP â€“ PR < OR or OP â€“ OQ' â‰¤ OR $\Rightarrow | z_{1} | - | z_{2} | \leq | z_{1} - z_{2} | \Rightarrow | z_{1} - z_{2} | \geq | z_{1} | - | z_{2} |$

prove the triangle inequality: IZ1 - z2I <or equal to Iz1I+Iz2I