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Question Number 19732 by Tinkutara last updated on 15/Aug/17

$$\mathrm{The}\mathrm{locus}\mathrm{of}z\mathrm{given}\mathrm{by}\mid \frac{z-1}{z-i}\mid =1\mathrm{is}$$

Answered by ajfour last updated on 15/Aug/17

$$\mathrm{z}\mathrm{lies}\mathrm{on}\mathrm{perpendicular}\mathrm{bisector}$$$$\mathrm{of}\mathrm{line}\mathrm{joining}\mathrm{z}=1\mathrm{and}\mathrm{z}=\mathrm{i}\mathrm{that}$$$$\mathrm{passes}\mathrm{through}\mathrm{origin}.$$$$\mathrm{So},\mathrm{y}=\mathrm{x}$$$$\frac{\mathrm{z}-\overline{\mathrm{z}}}{2\mathrm{i}}=\frac{\mathrm{z}+\overline{\mathrm{z}}}{2}\mathrm{or}\mathrm{z}-\overline{\mathrm{z}}=\mathrm{iz}+\mathrm{i}\overline{\mathrm{z}}$$$$\mathrm{or}(1-\mathrm{i})\mathrm{z}-(1+\mathrm{i})\overline{\mathrm{z}}=0$$$$\mathrm{multipying}\mathrm{by}\mathrm{i}\mathrm{we}\mathrm{get}$$$$(1+\mathrm{i})\mathrm{z}+(1-\mathrm{i})\overline{\mathrm{z}}=0.$$

Commented by Tinkutara last updated on 15/Aug/17

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

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