Mathematics - Complex Number Question with Solution | TestHub

Real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is given by $$\begin{array}{l}\frac{1}{2}\left[\frac{\bar{z}+2}{\bar{z}-1}+\left(\frac{\bar{z}+2}{\bar{z}-1}\right)\right]=4\\ \Rightarrow \frac{\bar{z}+2}{\bar{z}-1}+\frac{z+2}{z-1}=8\\ \Rightarrow z\bar{z}-\bar{z}+2z-2+z\bar{z}+2\bar{z}-z-2\\ =8(z\bar{z}-\bar{z}-z+1)\\ \Rightarrow z\bar{z}-\frac{3}{2}z-\frac{3}{2}\bar{z}+2=0\end{array}$$ Comparing with the equation $z\bar{z}+\bar{a}z+a\bar{z}+b=0,$ we get $a=-\frac{3}{2}$ and $b$ $=2$. Thus, the locus of $z$ given by the equation (i) is a circle with centre $=\frac{1}{2}$