Different ways of solving this simple modulus inequality

Question

Consider the set $S$ of all $z\in\mathbb C$ for which \begin{equation}\tag{1}\label{1} \left|\frac{2+z}{2-z}\right| \le 1. \end{equation}

One can easily find that $S=\{z\in\mathbb C \, :\, \Re(z) \le 0\}$ by using basic properties of the complex numbers (as Martin did; $\Re(z)$ denotes the real part of $z$).

What are some other interesting ways to arrive at this result?

Answer 1

Ok, it is for $$z\ne 2$$ equivalent to $$|2+x+iy|\le |2-x-iy|$$ so we obtain $$\sqrt{(2+x)^2+y^2}\le\sqrt{(2-x)^2+y^2}$$

Answer 2

Using the identities $|w|^2 = w \overline w$ and $2 \operatorname{Re}(w) = w + \overline w$ the inequality is equivalent to $$ |2+z|^2 \le |2-z|^2 \\ \iff 4 + 2z + 2\overline z + |z|^2 \le 4 - 2z - 2\overline z + |z|^2 \\ \iff 8 \cdot \operatorname{Re}(z) \le 0 $$

Answer 3

It's equivalent to $|2+z|\le|2-z|$ ($z=2$ is not possible in either case). Geometrically, this means the distance from the point $z$ to $-2$ is less than the distance from $z$ to $2$. You will get everything on one side of the perpendicular bisector of $2$ and $-2$, that is a half plane.

Note that if you replace "$\le1$" by "$\le k$" for some real $k>0$, you get (everything on one side of) a branch of hyperbola.