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Question Number 19739 by Tinkutara last updated on 15/Aug/17

$$\mathrm{If}z=x+iy\mathrm{is}\mathrm{a}\mathrm{complex}\mathrm{number}$$$$\mathrm{satisfying}\mid z+\frac{i}{2}{\mid }^2=\mid z-\frac{i}{2}{\mid }^2,\mathrm{then}$$$$\mathrm{the}\mathrm{locus}\mathrm{of}z\mathrm{is}$$

Answered by ajfour last updated on 15/Aug/17

$$\mathrm{y}=0\left(\mathrm{real}\mathrm{axis}\right)$$$$(\mathrm{z}+\frac{\mathrm{i}}{2})(\overline{\mathrm{z}}-\frac{\mathrm{i}}{2})=(\mathrm{z}-\frac{\mathrm{i}}{2})(\overline{\mathrm{z}}+\frac{\mathrm{i}}{2})$$$$\Rightarrow \mathrm{z}\overline{\mathrm{z}}-\frac{\mathrm{iz}}{2}+\frac{\mathrm{i}\overline{\mathrm{z}}}{2}+\frac{1}{4}=\mathrm{z}\overline{\mathrm{z}}+\frac{\mathrm{iz}}{2}-\frac{\mathrm{i}\overline{\mathrm{z}}}{2}+\frac{1}{4}$$$$\Rightarrow \mathrm{z}-\overline{\mathrm{z}}=0$$$$\mathrm{y}=0.$$

Commented by Tinkutara last updated on 15/Aug/17

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

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