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Question Number 21109 by Tinkutara last updated on 13/Sep/17

$$\mathrm{STATEMENT}-1:\mathrm{The}\mathrm{locus}\mathrm{of}z,\mathrm{if}$$$$\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi }{2}\mathrm{is}\mathrm{a}\mathrm{circle}.$$$$\mathbf{and}$$$$\mathrm{STATEMENT}-2:\mid \frac{z-2}{z+2}\mid =\frac{\pi }{2},\mathrm{then}$$$$\mathrm{the}\mathrm{locus}\mathrm{of}z\mathrm{is}\mathrm{a}\mathrm{circle}.$$

Commented by youssoufab last updated on 13/Sep/17

$$\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi }{2}$$$$\iff \arg \left(\frac{z-1}{z+1}\right)=\arg \left(i\right)$$$$\iff \frac{z-1}{z+1}\equiv i\left[2\pi \right]$$$$\iff z-1\equiv (z+1)i\left[2\pi \right]$$$$\iff z-1\equiv zi+i\left[2\pi \right]$$$$\iff z(1-i)\equiv i+1\left[2\pi \right]$$$$\iff z\equiv \frac{i+1}{1-i}\left[2\pi \right]$$$$\iff z\equiv \frac{{(i+1)}^2}{(1-i)(i+1)}\left[2\pi \right]$$$$\iff z\equiv \frac{-1+2i+1}{1+1}\left[2\pi \right]$$$$\iff z\equiv i\left[2\pi \right]$$$$z\mathrm{is}\mathrm{a}\mathrm{cercle}\Rightarrow z=i$$$$$$

Commented by Tinkutara last updated on 13/Sep/17

$$\mathrm{Locus}\mathrm{of}z\mathrm{will}\mathrm{not}\mathrm{be}\mathrm{a}\mathrm{complete}\mathrm{circle},$$$$\mathrm{but}\mathrm{why}\mathrm{it}\mathrm{is}\mathrm{not},\mathrm{is}\mathrm{my}\mathrm{doubt}.$$

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