1-iz-0-1-iz-0-show-that-z-0-R-

Question Number 127572 by mathocean1 last updated on 30/Dec/20

∣ 1 + {i z}_{0} ∣=∣ 1 - {i z}_{0} ∣

s h o w t h a t z_{0} \in R

Answered by mathmax by abdo last updated on 31/Dec/20

let z_{0} = x + iy so ∣ 1 + {iz}_{0} ∣=∣ 1 - {iz}_{0} ∣ \Rightarrow∣ 1 + i (x + iy) ∣=∣ 1 - i (x + iy) ∣ \Rightarrow

∣ 1 + ix - y ∣=∣ 1 - ix + y ∣ \Rightarrow \sqrt{{(1 - y)}^{2} + x^{2}} = \sqrt{{(1 + y)}^{2} + x^{2}} \Rightarrow

{(1 - y)}^{2} + x^{2} = {(1 + y)}^{2} + x^{2} \Rightarrow 1 - 2 y + y^{2} = 1 + 2 y + y^{2} \Rightarrow 4 y = 0 \Rightarrow y = 0 \Rightarrow

z_{0} = x \in R