Theorem:
Let $w, z \in \mathbb{C}$ such that $\bar{z}w \neq 1$. Then$$ |z| < 1 \land |w| < 1 \Rightarrow \left| \frac{w-z}{1-\bar{w}z}\right| < 1$$
I tried to make $w = u +iv, z = x + iy$ and expand, but that did not give me useful results. How to prove it?
$\endgroup$ 01 Answer
$\begingroup$Equivalently $$|w-z|^2<|1-\bar{w}z|^2$$ i.e. $$(w-z)(\bar{w}-\bar{z})<(1-\bar{w}z)(1-w\bar{z})$$ which is also equivalent to $$|w|^2+|z|^2<1+|w|^2|z|^2$$ The above inequality can be written as $$|w|^2(1-|z|^2)<1-|z|^2$$ which is always true for $|w|<1$, $|z|<1$.
$\endgroup$ 2
