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.

Let $z = x + iy$, then $|z+1| = \sqrt{(x+1)^2 + y^2}$ and $|z-1| = \sqrt{(x-1)^2 + y^2}$.
$|z+1| = 4 - |z-1|$
$(x+1)^2 + y^2 = 16 + (x-1)^2 + y^2 - 8\sqrt{(x-1)^2 + y^2}$
$x - 4 = -2\sqrt{(x-1)^2 + y^2}$
$x^2 + 16 - 8x = 4(x-1)^2 + 4y^2$
$12 = 4y^2 + 3x^2$
$\frac{x^2}{4} + \frac{y^2}{3} = 1$ which is formula of ellipse.
