Let $g(z) = \frac{z-a}{1-\overline{a}z}$ for $|a| < 1$. Prove that $|g(z)| = 1$ for $|z| = 1$

Question

Let $g(z) = \frac{z-a}{1-\overline{a}z}$ for $|a| < 1$. Prove that $|g(z)| = 1$ for $|z| = 1$. Note: $z,a \in \mathbb{C}$

So far I let $z=z_1+z_2i$ and $a=a_1+a_2i$ and then

$$g(z) = \frac{z-a}{1-\overline{a}z} = \frac{z_1+z_2i-a_1-a_2i}{1-(a_1-a_2i)(z_1+z_2i)}$$

$$= \frac{z_1-a_1+(z_2-a_2)i}{1-a_1z_1+a_2z_2+i(a_1z_2-z_1a_2)}$$

and then I took the modulus

$$|g(z)| = \sqrt{\frac{(z_1-a_1)^2+(z_2-a_2)^2}{(1-a_1z_1+a_2z_2)^2+(a_1z_2-z_1a_2)^2}}$$

But this approach is getting incredibly messy. Is there an easier and cleaner way to approach this problem?

Answer 1

Because $|1-a\overline z|=|z||1-a\overline z|=|z-az\overline z|=|z-a|$!