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

The locus of z given by ∣((z − 1)/(z − i))∣ = 1 is

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

Answered by ajfour last updated on 15/Aug/17

z lies on perpendicular bisector of line joining z=1 and z=i that passes through origin. So, y=x ((z−z^� )/(2i))=((z+z^� )/2) or z−z^� =iz+iz^� or (1−i)z−(1+i)z^� =0 multipying by i we get (1+i)z+(1−i)z^� =0 .

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

Commented by Tinkutara last updated on 15/Aug/17

Thank you very much Sir!

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

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