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Question Number 26374 by tawa tawa last updated on 24/Dec/17

$$\mathrm{show}\mathrm{that}\mathrm{if}\arg \left(\frac{{\mathrm{z}}_1+{\mathrm{z}}_2}{{\mathrm{z}}_1-{\mathrm{z}}_2}\right)=\frac{\pi }{2}\mathrm{then}\mid {\mathrm{z}}_1\mid =\mid {\mathrm{z}}_2\mid$$

Answered by Rasheed.Sindhi last updated on 31/Dec/17

$$\arg \left(\frac{{\mathrm{z}}_1+{\mathrm{z}}_2}{{\mathrm{z}}_1-{\mathrm{z}}_2}\right)=\frac{\pi }{2}$$$${\tan }^{-1}\frac{\mathrm{Im}\left(\frac{{\mathrm{z}}_1+{\mathrm{z}}_2}{{\mathrm{z}}_1-{\mathrm{z}}_2}\right)}{\mathrm{Re}\left(\frac{{\mathrm{z}}_1+{\mathrm{z}}_2}{{\mathrm{z}}_1-{\mathrm{z}}_2}\right)}=\frac{\pi }{2}$$$$\frac{\mathrm{Im}\left(\frac{{\mathrm{z}}_1+{\mathrm{z}}_2}{{\mathrm{z}}_1-{\mathrm{z}}_2}\right)}{\mathrm{Re}\left(\frac{{\mathrm{z}}_1+{\mathrm{z}}_2}{{\mathrm{z}}_1-{\mathrm{z}}_2}\right)}=\tan \left(\frac{\pi }{2}\right)=\mathrm{\infty }$$$$\therefore \mathrm{Re}\left(\frac{{\mathrm{z}}_1+{\mathrm{z}}_2}{{\mathrm{z}}_1-{\mathrm{z}}_2}\right)=0$$$$\mathrm{Let}{\mathrm{z}}_1={\mathrm{x}}_1+{\mathrm{iy}}_1\mathrm{\&}{\mathrm{z}}_2={\mathrm{x}}_2+{\mathrm{iy}}_2$$$$\mathrm{Re}\left(\frac{({\mathrm{x}}_1+{\mathrm{iy}}_1)+({\mathrm{x}}_2+{\mathrm{iy}}_2)}{({\mathrm{x}}_1+{\mathrm{iy}}_1)-({\mathrm{x}}_2+{\mathrm{iy}}_2)}\right)$$$$\mathrm{Re}\left(\frac{({\mathrm{x}}_1+{\mathrm{x}}_2)+\mathrm{i}({\mathrm{y}}_1+{\mathrm{y}}_2)}{({\mathrm{x}}_1-{\mathrm{x}}_2)+\mathrm{i}({\mathrm{y}}_1-{\mathrm{y}}_2)}\right)$$$$\mathrm{Let}{\mathrm{x}}_1+{\mathrm{x}}_2={\mathrm{u}}_1,{\mathrm{y}}_1+{\mathrm{y}}_2={\mathrm{v}}_1$$$${\mathrm{x}}_1-{\mathrm{x}}_2={\mathrm{u}}_2\mathrm{and}{\mathrm{y}}_1-{\mathrm{y}}_2={\mathrm{v}}_2$$$$\mathrm{Re}\left(\frac{{\mathrm{u}}_1+{\mathrm{iv}}_1}{{\mathrm{u}}_2+{\mathrm{iv}}_2}\right)=\mathrm{Re}\left(\frac{{\mathrm{u}}_1+{\mathrm{iv}}_1}{{\mathrm{u}}_2+{\mathrm{iv}}_2}\times \frac{{\mathrm{u}}_2-{\mathrm{iv}}_2}{{\mathrm{u}}_2-{\mathrm{iv}}_2}\right)$$$$=\mathrm{Re}\left(\frac{{\mathrm{u}}_1{\mathrm{u}}_2+{\mathrm{v}}_1{\mathrm{v}}_2+\mathrm{i}({\mathrm{u}}_2{\mathrm{v}}_1-{\mathrm{u}}_1{\mathrm{v}}_2)}{{\mathrm{u}}_2^2+{\mathrm{v}}_2^2}\right)$$$$=\frac{{\mathrm{u}}_1{\mathrm{u}}_2+{\mathrm{v}}_1{\mathrm{v}}_2}{{\mathrm{u}}_2^2+{\mathrm{v}}_2^2}=0$$$${\mathrm{u}}_1{\mathrm{u}}_2+{\mathrm{v}}_1{\mathrm{v}}_2=0$$$$({\mathrm{x}}_1+{\mathrm{x}}_2)({\mathrm{x}}_1-{\mathrm{x}}_2)+({\mathrm{y}}_1+{\mathrm{y}}_2)({\mathrm{y}}_1-{\mathrm{y}}_2)=0$$$$({\mathrm{x}}_1^2-{\mathrm{x}}_2^2)+({\mathrm{y}}_1^2-{\mathrm{y}}_2^2)=0$$$$({\mathrm{x}}_1^2+{\mathrm{y}}_1^2)-({\mathrm{x}}_2^2+{\mathrm{y}}_2^2)=0$$$$({\mathrm{x}}_1^2+{\mathrm{y}}_1^2)=({\mathrm{x}}_2^2+{\mathrm{y}}_2^2)$$$$\mid {\mathrm{z}}_1{\mid }^2=\mid {\mathrm{z}}_2{\mid }^2$$$$\mid {\mathrm{z}}_1\mid =\mid {\mathrm{z}}_2\mid$$

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