Take $z_1,z_2 \in \mathbb{C},$ and assume $|z_1| < 1, |z_2| < 1.$ I want to prove that $$\left|\frac{z_1-z_2}{1-\bar{z}_1z_2}\right| < 1.$$ Take $z_1 = x+iy, z_2 = a+ib$ for $x,y,a,b \in \mathbb{R}.$ We see that $$|z_1| < 1 \implies \sqrt{x^2 + y^2} < 1 \implies x^2 + y^2 < 1.$$ similarly we see that $$|z_2| < 1 \implies \sqrt{a^2 + b^2} < 1 \implies a^2 + b^2 < 1.$$ We see that $$\left|\frac{z_1-z_2}{1-\bar{z}_1z_2}\right| =\left|\frac{(x-a) + i(y-b)}{1 -(x-iy)(a+ib)}\right| =\left|\frac{(x-a) + i(y-b)}{1 -(xa + ibx -iay + yb)}\right| =\left|\frac{(x-a) + i(y-b)}{(1-xa-yb) + i(ay - bx)}\right|.$$ However, I am beginning to have some issues as to how to simplify this to a non-fraction. Any recommendations?
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Handling with complex numbers directly would be much easier. Take the square on both side of the desired inequality and use the formula $$ |z|^2=z\bar z. $$
Eventually, you end up with $$ |a|^2+|b|^2-\bar ab-a\bar b<1-\bar ab-a\bar b-|a|^2|b|^2. $$ Now use the assumption $|a|<1$ and $|b|<1$.
The transformation in your question is called Blaschke Factor.