Finding which complex numbers satisfy $\left|z + 1\right| + \left|z - 1\right| = 4$ This problem is from the book Complex Analysis with Applications by Asmar. Currently I am stuck at, well, everything, as I don't know know how to manipulate the expression $\left|z + 1\right| + \left|z - 1\right| = 4$ into anything meaningful. Squaring and simplifying both sides yields $z\bar{z} + \left|z + 1\right|\left|z - 1\right| + 1 = 8$, and I don't see how that is useful.
The problem is asking us to find those complex numbers whose sum of distances from both $1$ and $-1$ is equal to $4$. I suppose the end result is a circle of some kind, as points $(2, 0)$ and $(-2, 0)$ satisfy the equality.

If you write $z$ as $a+bi$ with $a,b\in\Bbb R$, then\begin{align}|z+1|+|z-1|=4&\iff\sqrt{(a+1)^2+b^2}+\sqrt{(a-1)^2+b^2}=4\\&\iff2a^2+2b^2+2+2\sqrt{(a+1)^2+b^2}\sqrt{(a-1)^2+b^2}=16\\&\iff\sqrt{(a+1)^2+b^2}\sqrt{(a-1)^2+b^2}=7-a^2-b^2\\&\iff\bigl((a+1)^2+b^2\bigr)\bigl((a-1)^2+b^2\bigr)=(7-a^2-b^2)^2\\&\iff3a^2+4b^2=12.\end{align}So, you get the ellipse $3a^2+4b^2=12$.
However, note that the equivalence\begin{multline}\sqrt{(a+1)^2+b^2}\sqrt{(a-1)^2+b^2}=7-a^2-b^2\iff\\\iff\bigl((a+1)^2+b^2\bigr)\bigl((a-1)^2+b^2\bigr)=(7-a^2-b^2)^2\end{multline}is not obvious, but is not hard to prove that $|z-1|+|z+1|=4\implies|z|^2\leqslant7$.
