Consider the following Mobius transformation: $$g(z)=\frac{az-\bar{c}}{cz+\bar{a}}$$ such that $$|a|^2+|c|^2=1$$ and the following metric $$d(z,w)=\frac{2|z-w|}{\sqrt{1+|z|^2}\sqrt{1+|w|^2}}.$$
From this, I need to show that distance is preserved. In particular, I'm stuck on the following step:
$$\sqrt{1+|g(z)|^2}=\sqrt{1+g(z)\overline{g(z)}}=\sqrt{1+\frac{az-\overline{c}}{cz+\overline{a}}\frac{\overline{a}\overline{z}-c}{\overline{c}\overline{z}+a}}$$
Just looking at the right hand side of what's under the square root:
$$(az-\overline{c})(\overline{a}\overline{z}-c)=a\overline{a}z\overline{z}-acz-\overline{a}\overline{c}\overline{z}+\overline{c}c$$
$$(cz+\overline{a})(\overline{cz}+a)=c\overline{c}z\overline{z}+acz+\overline{a}\overline{c}\overline{z}+a\overline{a}$$
Putting this together:
$$1+\frac{a\overline{a}z\overline{z}-acz-\overline{a}\overline{c}\overline{z}+\overline{c}c}{c\overline{c}z\overline{z}+acz+\overline{a}\overline{c}\overline{z}+a\overline{a}}$$
I see a few places for simplification, such as $a\overline{a}=|a|^2$:
$$1+\frac{|a|^2|z|^2-acz-\overline{acz}+|c|^2}{|c|^2|z|^2+acz+\overline{acz}+|a|^2}$$
$$\frac{|c|^2|z|^2+acz+\overline{acz}+|a|^2+(|a|^2|z|^2-acz-\overline{acz}+|c|^2)}{|c|^2|z|^2+acz+\overline{acz}+|a|^2}$$
This further reduces to: $$\frac{|c|^2|z|^2+|a|^2+|a|^2|z|^2+|c|^2}{|c|^2|z|^2+acz+\overline{acz}+|a|^2}=\frac{|z|^2+1}{|c|^2|z|^2+acz+\overline{acz}+|a|^2}$$
That's where I'm stuck trying to clean it up.
Hint:) Just simple calculations: $$1+|g(z)|^2= \frac{|c|^2|z|^2+|a|^2+|a|^2|z|^2+|c|^2}{|c|^2|z|^2+acz+\overline{acz}+|a|^2}=\frac{(|z|^2+1)(|a|^2+|c|^2)}{(cz+\overline{a})(\overline{c}\overline{z}+a)}$$ $$1+|g(w)|^2=\frac{(|w|^2+1)(|a|^2+|c|^2)}{(cw+\overline{a})(\overline{c}\overline{w}+a)}$$ $$|g(z)-g(w)|=\frac{|z-w|(|a|^2+|c|^2)}{(cz+\overline{a})(cw+\overline{a})}$$