Question

If the real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is $4,z\ne 1$ , then the locus of the point representing $z$ in the complex plane is

• A. a straight line parallel to x-axis
• B. a straight line equally inclined to axes
• C. a circle with radius 2
• D. a circle with radius $\frac{1}{2}$

Answer 1

Answer: (D)

Real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is given by
$\frac{1}{2}\left[\frac{\bar{z}+2}{\bar{z}-1}+\left(\frac{\bar{z}+2}{\bar{z}-1}\right)\right]=4$
$\Rightarrow \frac{\bar{z}+2}{\bar{z}-1}+\frac{z+2}{z-1}=8$
$\Rightarrow z\bar{z}-\bar{z}+2z-2+z\bar{z}+2\bar{z}-z-2$
$=8(z\bar{z}-\bar{z}-z+1)$
$\Rightarrow z\bar{z}-\frac{3}{2}z-\frac{3}{2}\bar{z}+2=0)$ ...(i)
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 center
$\frac{3}{2}$ and radius $=\frac{1}{2}$