$P$ represents the variable complex number $z$. Find the locus of $P$, if $\;$ $arg \left(\dfrac{z - 1}{z + 3}\right) = \dfrac{\pi}{4}$
Let $z = x + iy$
Given: $\;\;\;$ $arg \left(\dfrac{z - 1}{z + 3}\right) = \dfrac{\pi}{4}$
i.e. $\;$ $arg \left(z - 1\right) - arg \left(z + 3\right) = \dfrac{\pi}{4}$
i.e. $\;$ $arg \left[\left(x - 1\right) + i y\right] - arg \left[\left(x + 3\right) + i y\right] = \dfrac{\pi}{4}$
i.e. $\;$ $\tan^{-1} \left(\dfrac{y}{x - 1}\right) - \tan^{-1} \left(\dfrac{y}{x + 3}\right) = \dfrac{\pi}{4}$
i.e. $\;$ $\tan^{-1} \left[\dfrac{\dfrac{y}{x - 1} - \dfrac{y}{x + 3}}{1 + \left(\dfrac{y}{x - 1}\right) \left(\dfrac{y}{x + 3}\right)} \right] = \dfrac{\pi}{4}$
i.e. $\;$ $\tan^{-1} \left[\dfrac{4y}{x^2 + 2x - 3 + y^2}\right] = \dfrac{\pi}{4}$
i.e. $\;$ $\dfrac{4y}{x^2 + y^2 + 2x - 3} = \tan \left(\dfrac{\pi}{4}\right) = 1$
i.e. $\;$ $x^2 + y^2 + 2x - 4y - 3 = 0$ $\;\;\; \cdots \; (1)$
Equation $(1)$ is the required equation of locus of $P$.