Solve $| \frac{2+z}{2-z} | < 1$

Question

How do I solve this equation $$\left| \frac{2+z}{2-z} \right| < 1$$ for complex $z \in \Bbb C$?

I know the answer is $\text{Re}(z) \lt 0$ but I can not understand how to get there. I tried making the denominator real but how do I proceed from here?

$$\left| \frac{2+z}{2-z} \right| = \left| \frac{(2+z)^2}{(2-z)(2+z)} \right| = \left| \frac{4+4z+z^2}{4 - z^2} \right|$$

Answer 1

Rewrite like this $$|z-(-2)|<|z-2|$$ so $z$ is closer to $-2$ than to $2$. The set of points $z$ such that $z$ is equally apart from $2$ and $-2$ is $ \operatorname{Re}(z)=0$ so in this case we have $ \operatorname{Re}(z)<0$.

Answer 2

A more laborious algebraic approach is given here. The condition $\left|\frac{2+z}{2-z}\right|<1$ is equivalent to $$\begin{align}4+4\,\text{Re}(z)+|z|^2&=4+2z+2\bar{z}+z\bar{z}=(2+z)(2+\bar{z})=|2+z|^2 \\&<|2-z|^2=(2-z)(2-\bar{z})=4-2z-2\bar{z}+z\bar{z}=4-4\,\text{Re}(z)+|z|^2\,.\end{align}$$ Thus, $\left|\frac{2+z}{2-z}\right|<1$ if and only if $\text{Re}(z)<0$.