Q44 / 125 · BITSAT 2013 · Mathematics
If the real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is $4, z \neq 1,$ then the locus of the point representing $z$ in the complex plane is
- Aa straight line parallel to $x$ -axis
- Ba straight line equally inclined to axes
- Ca circle with radius 2
- Correct answerDa circle with radius $\frac{1}{2}$
Solution
Real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is given by
$$
\begin{array}{l}
\frac{1}{2}\left[\frac{\bar{z}+2}{\bar{z}-1}+\left(\frac{\bar{z}+2}{\bar{z}-1}\right)\right]=4 \\
\Rightarrow \frac{\bar{z}+2}{\bar{z}-1}+\frac{z+2}{z-1}=8 \\
\Rightarrow z \bar{z}-\bar{z}+2 z-2+z \bar{z}+2 \bar{z}-z-2 \\
=8(z \bar{z}-\bar{z}-z+1) \\
\Rightarrow z \bar{z}-\frac{3}{2} z-\frac{3}{2} \bar{z}+2=0
\end{array}
$$
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 $=\frac{1}{2}$
Hence, the correct answer is option: 'D'.