Explanation

(d)
$\frac{z - 1}{2z + 1} = \frac{(x - 1) + iy}{(2x + 1) + 2iy} \times \frac{(2x + 1) - 2iy}{(2x + 1) - 2iy}$
$= \frac{\begin{array}{r}
\left\{ (x + 1)(2x + 1) + 2y^{2} \right\} + \\
iy\{ - 2x + 2 + 2x + 1\}
\end{array}}{(2x + 1)^{2} + 4y^{2}}$
Given, $Im\left( \frac{z - 1}{2z + 1} \right) = - 4$
$\therefore\ \ \ \ \ \ \frac{3y}{(2x + 1)^{2} + 4y^{2}} = - 4$
$\Rightarrow \ 16x^{2} + 16y^{2} + 16x + 3y + 4 = 0$
$\therefore$ The locus of $z$ is a circle.