Small inequality on unit open disc For $|u|,|z|<1$, $u,z$ complex numbers, how to show the inequality:
$|\frac{u-z}{1-\bar uz}|<1$?

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 $<1$
