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Question Number 19350 by Tinkutara last updated on 10/Aug/17

$$\mathrm{Prove}\mathrm{that}\mid z_1+z_2\mid =\mid z_1-z_2\mid \iff$$$$\arg \left(z_1\right)-\arg \left(z_2\right)=\frac{\pi }{2}$$

Answered by ajfour last updated on 10/Aug/17

$$\Rightarrow \mid {\mathrm{z}}_1+{\mathrm{z}}_2{\mid }^2=\mid {\mathrm{z}}_1-{\mathrm{z}}_2{\mid }^2$$$${\mathrm{z}}_1^2+{\mathrm{z}}_2^2+2\mathrm{Re}\left({\mathrm{z}}_1{\overline{\mathrm{z}}}_2\right)={\mathrm{z}}_1^2+{\mathrm{z}}_2^2-2\mathrm{Re}\left({\mathrm{z}}_1{\overline{\mathrm{z}}}_2\right)$$$$\Rightarrow \mathrm{Re}\left({\mathrm{z}}_1{\overline{\mathrm{z}}}_2\right)=0$$$$\mathrm{Re}\left[({\mathrm{x}}_1+{\mathrm{iy}}_1)({\mathrm{x}}_2-{\mathrm{iy}}_2)\right]=0$$$$\Rightarrow {\mathrm{x}}_1{\mathrm{x}}_2=-{\mathrm{y}}_1{\mathrm{y}}_2$$$$\Rightarrow \frac{{\mathrm{y}}_2}{{\mathrm{x}}_2}=-\frac{{\mathrm{x}}_1}{{\mathrm{y}}_1}$$$$\Rightarrow \tan \left[\arg \left({\mathrm{z}}_2\right)\right]=-\frac{1}{\tan \left[\arg \left({\mathrm{z}}_1\right)\right]}$$$$\Rightarrow \tan {\theta }_1\tan {\theta }_2+1=0$$$$\mathrm{and}\mathrm{as}\tan ({\theta }_1-{\theta }_2)=\frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}$$$$\Rightarrow {\theta }_1-{\theta }_2=\frac{\pi }{2}$$$$\arg \left({\mathrm{z}}_1\right)-\arg \left({\mathrm{z}}_2\right)=\frac{\pi }{2}.$$

Commented by Tinkutara last updated on 10/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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