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

$$\mid 1+{iz}_0\mid =\mid 1-{iz}_0\mid$$$$showthat{z}_0\in \mathbb{R}$$

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

$$\mathrm{let}{\mathrm{z}}_0=\mathrm{x}+\mathrm{iy}\mathrm{so}\mid 1+{\mathrm{iz}}_0\mid =\mid 1-{\mathrm{iz}}_0\mid \Rightarrow \mid 1+\mathrm{i}(\mathrm{x}+\mathrm{iy})\mid =\mid 1-\mathrm{i}(\mathrm{x}+\mathrm{iy})\mid \Rightarrow$$$$\mid 1+\mathrm{ix}-\mathrm{y}\mid =\mid 1-\mathrm{ix}+\mathrm{y}\mid \Rightarrow \sqrt{{(1-\mathrm{y})}^2+{\mathrm{x}}^2}=\sqrt{{(1+\mathrm{y})}^2+{\mathrm{x}}^2}\Rightarrow$$$${(1-\mathrm{y})}^2+{\mathrm{x}}^2={(1+\mathrm{y})}^2+{\mathrm{x}}^2\Rightarrow 1-2\mathrm{y}+{\mathrm{y}}^2=1+2\mathrm{y}+{\mathrm{y}}^2\Rightarrow 4\mathrm{y}=0\Rightarrow \mathrm{y}=0\Rightarrow$$$${\mathrm{z}}_0=\mathrm{x}\in \mathrm{R}$$

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