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Question Number 19734 by Tinkutara last updated on 15/Aug/17

$$\mathrm{If}\mathrm{the}\mathrm{imaginary}\mathrm{part}\mathrm{of}\frac{2z+1}{iz+1}\mathrm{is}-2,$$$$\mathrm{then}\mathrm{the}\mathrm{locus}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{representing}$$$$z\mathrm{in}\mathrm{the}\mathrm{complex}\mathrm{plane}\mathrm{is}$$

Answered by ajfour last updated on 15/Aug/17

$$\frac{2\mathrm{z}+1}{\mathrm{iz}+1}=\frac{(2\mathrm{z}+1)(1-\mathrm{i}\overline{\mathrm{z}})}{(\mathrm{iz}+1)(1-\mathrm{i}\overline{\mathrm{z}})}$$$$=\frac{2\mathrm{z}-2\mathrm{i}\mid {\mathrm{z}}^2\mid +1-\mathrm{i}\overline{\mathrm{z}}}{\mathrm{iz}+\mid \mathrm{z}{\mid }^2+1-\mathrm{i}\overline{\mathrm{z}}}$$$$\mathrm{if}\mathrm{z}=\mathrm{x}+\mathrm{iy}$$$$=\frac{2(\mathrm{x}+\mathrm{iy})-2\mathrm{i}({\mathrm{x}}^2+{\mathrm{y}}^2)+1-\mathrm{ix}-\mathrm{y}}{1+{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{y}}$$$$\mathrm{as}\mathrm{imaginary}\mathrm{part}\mathrm{of}\mathrm{above}\mathrm{complex}$$$$\mathrm{number}\mathrm{is}=-2,\mathrm{we}\mathrm{can}\mathrm{write}$$$$\frac{2\mathrm{y}-{2\mathrm{x}}^2-{2\mathrm{y}}^2-\mathrm{x}}{1+{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{y}}=-2$$$$\mathrm{or}2\mathrm{y}+\mathrm{x}-2=0$$$$\Rightarrow 4\mathrm{iy}+2\mathrm{ix}-4\mathrm{i}=0$$$$\mathrm{or}2(\mathrm{z}-\overline{\mathrm{z}})+\mathrm{i}(\mathrm{z}+\overline{\mathrm{z}})-4\mathrm{i}=0$$$$(2+\mathrm{i})\mathrm{z}-(2-\mathrm{i})\overline{\mathrm{z}}-4\mathrm{i}=0$$$$\Rightarrow (1-2\mathrm{i})\mathrm{z}+(1+2\mathrm{i})\overline{\mathrm{z}}+4=0.$$

Commented by Tinkutara last updated on 15/Aug/17

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

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