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Question Number 33095 by 7991 last updated on 10/Apr/18

$$z=x+yi\in \mathbb{C}$$$$\stackrel{-}{z}=x-y\in \mathbb{C}$$$$proof\mid z{\mid }^2=\mid z^2\mid =z\stackrel{-}{z},soz\ne 0\to \frac{1}{z}=\frac{\stackrel{-}{z}}{\mid z{\mid }^2}$$

Answered by Rasheed.Sindhi last updated on 10/Apr/18

$$z=x+yi\in \mathbb{C},\bar{z}=x-yi$$$$\mid \mathrm{z}\mid =\mid x+yi\mid =\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}$$$$\therefore \mid \mathrm{z}{\mid }^2={\mathrm{x}}^2+{\mathrm{y}}^2.............\mathrm{A}$$$$\mid {\mathrm{z}}^2\mid =\mid {(x+yi)}^2\mid =\mid {\mathrm{x}}^2-{\mathrm{y}}^2+2\mathrm{xyi}\mid$$$$=\sqrt{{({\mathrm{x}}^2-{\mathrm{y}}^2)}^2+{\left(2\mathrm{xy}\right)}^2}$$$$=\sqrt{{\mathrm{x}}^4+{\mathrm{y}}^4-{2\mathrm{x}}^2{\mathrm{y}}^2+{4\mathrm{x}}^2{\mathrm{y}}^2}$$$$=\sqrt{{\mathrm{x}}^4+{\mathrm{y}}^4+{2\mathrm{x}}^2{\mathrm{y}}^2}$$$$=\sqrt{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}$$$$={\mathrm{x}}^2+{\mathrm{y}}^2...............\mathrm{B}$$$$\mathrm{z}.\bar{\mathrm{z}}=(x+yi)(x-yi)={\left(\mathrm{x}\right)}^2-{\left(\mathrm{yi}\right)}^2$$$$={\mathrm{x}}^2+{\mathrm{y}}^2.................\mathrm{C}$$$$\mathrm{From}\mathrm{A},\mathrm{B}\mathrm{\&}\mathrm{C}$$$$\mid \mathrm{z}{\mid }^2=\mid {\mathrm{z}}^2\mid =\mathrm{z}.\bar{\mathrm{z}}$$$$\mathrm{z}.\bar{\mathrm{z}}=\mid {\mathrm{z}}^2\mid \Rightarrow \mathrm{z}=\frac{\mid {\mathrm{z}}^2\mid }{\stackrel{-}{\mathrm{z}}}\Rightarrow \frac{1}{\mathrm{z}}=\frac{\stackrel{-}{\mathrm{z}}}{\mid {\mathrm{z}}^2\mid }$$

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