If a complex number $z$ is such that $\frac{z-2 i}{z-2}$ purely imaginary number and the locus of $z$ is a closed curve, then the area of the region bounded by that closed curve and lying in the first quadrant is

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Aicharya · Accepted Answer

Given, $\frac{z-2 i}{z-2}$ is purely imaginary. $\begin{aligned} & \Rightarrow \frac{z-2 i}{z-2}+\left(\frac{\overline{z-2 i}}{z-2}ight)=2 \operatorname{Re}\left(\frac{z-2 i}{z-2}ight) \ & \Rightarrow \frac{z-2 i}{z-2}+\frac{\bar{z}+2 i}{\bar{z}-2}=0 \ & \Rightarrow|z|^2-2 z-2 i \bar{z}+4 i+|z|^2+2 i z-2 \bar{z}-4 i= \ & \Rightarrow 2|z|^2-2(z+\bar{z})+2 i(z-\bar{z})=0 \ & \Rightarrow x^2+y^2-2 x-2 y=0 \quad(\because z=x+i y) \ & \Rightarrow(x-1)^2+(y-1)^2=(\sqrt{2})^2 \ & \Rightarrow 	ext { Required area }=\frac{\pi r^2}{2}=\frac{\pi(\sqrt{2})^2}{4}=\frac{\pi}{2} \end{aligned}$

If a complex number $z$ is such that $\frac{z-2 i}{z-2}$ purely imaginary number and the locus of $z$ is a…

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