Different ways of solving this simple modulus inequality 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?

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