Knowing that $p$ and $q$ are complex numbers, $|p| < 1$ and $|q|<1$ show that $|\frac{p - q}{1 - q\bar{p}}| < 1$.
I've tried to write:
$p=x + yi$ and $q=a+bi$ which led me to $x^2 + y^2 + a^2 + b^2 < 1 + (x^2 + y^2)(a^2 + b^2)$ after some algebra, but I've not been able to continue from that point.
For an entirely different approach that doesn't involve wrangling with algebra, you can try defining a function via
$$f(z) = \frac{p - z}{1 - \overline{p} z}$$
If $1 - \overline{p} z = 0$, then $|z| = \frac{1}{|\overline{p}|} = \frac{1}{|p|} > 1$; hence, $f$ is analytic on a neighborhood of the closed unit disk.
On the other hand, if $|z| = 1$, we see that
$$|f(z)| = \frac{1}{|z|} \frac{|p - z|}{|1/z - \overline{p}|} = 1 \frac{|p - z|}{|\overline{z} - \overline{p}|} = 1$$
Hence, by the Maximum Modulus Principle, $|f(z)| \leq 1$ for all $|z| < 1$. If equality held for some $z$, then $f$ attains a maximum within the disk, and so is constant. This clearly does not hold, and the result follows.
You have already got it. $$|p|^2+|q|^2<1+|pq|^2 \implies(1-|p|^2)(1-|q|^2)>0$$ which is true!
