$P$ represents the variable complex number $z$.Find the locus of $P$, if $\;$ $Im \left[\dfrac{2z + 1}{iz + 1}\right] = -2$
Let $P$ be the point $z = x + iy$
Then,
$\begin{aligned}
\dfrac{2z + 1}{iz + 1} & = \dfrac{\left(2x + 1\right) + i \; 2y}{\left(1 - y\right) + i \; x} \\\\
& = \dfrac{\left[\left(2x + 1\right) + i \; 2y\right] \left[\left(1 - y\right) - i \;x\right]}{\left(1 -y\right)^2 - \left(i \; x\right)^2} \\\\
& = \dfrac{\left(2x + 1\right) \left(1 - y\right) - i \; x \left(2x + 1\right) + i \; 2y \left(1 - y\right) - i^2 \; 2xy}{1 -2y + y^2 - i^2 \; x^2} \\\\
& = \dfrac{\left(2x + 1\right) \left(1 - y\right) + 2xy}{x^2 + y^2 - 2y + 1} + i \;\dfrac{2y \left(1 - y\right) - x \left(2x + 1\right)}{x^2 + y^2 - 2y + 1}
\end{aligned}$
$\therefore$ $\;$ $Im \left[\dfrac{2z + 1}{iz + 1}\right] = \dfrac{2y - 2 y^2 -2 x^2 - x}{x^2 + y^2 - 2y + 1}$
Given $\;\;$ $\dfrac{2y - 2 y^2 -2 x^2 - x}{x^2 + y^2 - 2y + 1} = -2$
i.e. $\;$ $2 y - 2 y^2 - 2 x^2 - x = - 2 x^2 - 2 y^2 + 4 y - 2$
i.e. $\;$ $x + 2y - 2 = 0$ $\;\;\; \cdots \; (1)$
Equation $(1)$ is the required equation of locus of $P$.