Show inequality of complex number: $|\frac{a+b}{1+a\bar{b}}|<1$

Question

Suppose $a,b\in\mathbb{C},|a|<1,|b|<1$, how to see $\displaystyle\left|\frac{a+b}{1+a\bar{b}}\right|<1$?

Answer 1

If you write the inequality as $|a+b|^2 < |1+a \bar{b}|^2$ and use $|z|^2 = z \bar{z}$, then it is enough to show that $$|a|^2+|b|^2 < 1+|a|^2|b|^2$$ when $|a|<1$ and $|b|<1$. Now move the left terms to the right and factor to get $$0 < (1-|a|^2)(1-|b|^2)$$ which is clear, given the restrictions on $|a|$ and $|b|$.