Small inequality on unit open disc For $|u|,|z|<1$, $u,z$ complex numbers, how to show the inequality:
$|\frac{u-z}{1-\bar uz}|<1$?
 A: Let $u=a+ib, z=c+id$
So, $|u-z|=\sqrt{(a-c)^2+(b-d)^2}=\sqrt{a^2+c^2-2ca+b^2+d^2-2bd}$
$1-\bar uz=1-(a-ib)(c+id)=1-ac-bc+i(bc-ad)$
So, $|1-\bar uz|=\sqrt{(1-ac-bc)^2+(bc-ad)^2}$
$|u-v|<|1-\bar uz|$
$\implies \sqrt{a^2+c^2-2ca+b^2+d^2-2bd}<\sqrt{(1-ac-bc)^2+(bc-ad)^2}$
$\implies a^2+c^2-2ca+b^2+d^2-2bd <(1-ac-bc)^2+(bc-ad)^2$
$\implies \{1-(a^2+b^2)\}\{1-(c^2+d^2)\}<0$
$\implies \{1-|u|^2\}\{1-|z|^2\}<0$
which is satisfied if the modulus of $u,z$ both $>1$ or both $<1$
A: Suppose $u=re^{i\theta}$, we make the substitution
$$z\leftrightarrow ze^{i\theta}$$
and we find
$$|\frac{re^{i\theta}-ze^{i\theta}}{1-re^{-i\theta}ze^{i\theta}}|=|\frac{r-z}{1-rz}|$$
so we may assume that $u$ is a real number $0<r<1$. We want to show that
$$(r-z)(r-\overline{z})<(1-rz)(1-\overline{rz})$$
We expand both sides, make the necessary cancellations and move all terms to the right, to see that the above inequality is equivalent to
$$0<(1-r^2)(1-z\overline{z})$$
hence
$$0<(1-r^2)(1-|z|^2)$$
Now it is easy to conclude (and also to note that $=1$ in the assignment precisely holds when $|u|=1$ or $|z|=1$)
