Complex locus with sum of arguments

Question

Find the locus of $z$ when: $$\arg(z+2) + \arg(z-2) = \pi. $$

I tried substituting $ z = x + iy$ but the algebra is fairly messy and I'm not sure how to work through it. Is there a better approach to solving the question.

Thanks.

Answer 1

From $$\arg (z+2) + \arg (z-2) = \arg (z^2-4) = \pi $$

we deduce that $\;z^2-4\;$ is a negative real number. This occurs when

1. $z\in \mathbb{R},\; -2<z<2,$ or

2. $z=i\alpha,\; \alpha \in \mathbb{R},$ is pure imaginary.

Answer 2

HINT

A trivial case is when $z$ is real.

Excluding the trivial case, let consider different cases, for example for $z$ in the first quadrant we need that $z-2$ is in the second quadrant and

$$\arg(z+2) + \arg(z-2) = \pi \iff \arctan\left(\frac{y}{x+2}\right)+\arctan\left(\frac{y}{x-2}\right)+\pi=\pi$$