How to prove the modulus of $\frac{z-w}{z-\bar{w}} < 1$? Given that $\Im{(z)} > 0$ and $\Im{(w)} > 0$, prove that $|\displaystyle\frac{z-w}{z-\bar{w}}|<1$ . 
Please help me check my answer:
$z - w = a + ib$
$z - \bar{w}$ = $a + i(b+2\Im(w)) $
$|\displaystyle\frac{z-w}{z-\bar{w}}$| =  $\displaystyle\frac{|a+ib|}{|a+i(b+2\Im(w)|}$
= $\displaystyle\frac{\sqrt{(a^2+b^2)}}{\sqrt(a^2+(b+2\Im(w))^2)}$
< $\displaystyle\frac{\sqrt{(a^2+b^2)}}{\sqrt{(a^2+b^2)}}$ = 1

An example of almost pretty proof (In my eyes of course) using all the givens implicitly:
$$x \equiv z-w \implies y \equiv z - \overline{w}=x+2i\Im{(w)} \implies 0 <|x|^2 < |y|^2 \implies  |\frac{x}{y}|^2=\frac{|x|^2}{|y|^2}<1 \implies |\frac{z-w}{z - \overline{w}}|=|\frac{x}{y}|<1$$
