I have been stumped on this problem for a little bit, I figure that it's really simple and I'm missing something obvious, but I just wanted to see if what I have come up with is correct:
Let z and w be complex numbers such that |z|,|w|<1. Prove that: $|\dfrac{z-w}{1-\bar{z}w}|<1$.
Here's my proof: $|z-w|<1$, and $|\bar{z}w|=|\bar{z}||w|=|z||w|<1$, due to hypothesis. Therefore: $|z-w|-|\bar{z}||w|<1\Longrightarrow |z-w|-|\bar{z}w|<1\Longrightarrow |z-w|<1+|\bar{z}w|<|1-\bar{z}w|\Longrightarrow |\dfrac{z-w}{1-\bar{z}w}|<1$
$\square$
Is this correct, is there some loophole that I'm missing, or is there a simpler way?
Not sure how you know that $|z-w|<1$ to start with.
Hint. $$\eqalign{\left|\frac{z-w}{1-\overline zw}\right|<1\quad &\Leftrightarrow\quad |z-w|<|1-\overline zw|\cr &\Leftrightarrow\quad |z-w|^2<|1-\overline zw|^2\cr &\Leftrightarrow\quad (z-w)(\overline z-\overline w) <(1-\overline zw)(1-z\overline w)\ .\cr}$$