# If $g(a) = 0$, then $g$ takes unit disk to unit disk: conformal self map of unit disk

I'm trying to understand the proof of the theorem that the conformal self map of the unit disk $$U = \{ z \colon \left|z \right| < 1 \}$$ are exactly the fractional linear transformation of the form

$$f(z) = e^{i \phi} \frac{z-a}{1-\overline{a}z}$$ where $$\left|a \right| < 1$$ and $$0 \leq \phi \leq 2\pi$$

In the proof, to show that such an $$f$$ is indeed conformal self map, we first show that $$g(z) = \frac{z-a}{1-\overline{a}z}$$ takes the unit circle to unit circle and the proof says that as $$g(a) = 0$$, it must map unit disk to unit disk. This is the step that I do not understand. Why is it that $$g(a) = 0$$ implies $$g$$ maps unit disk to unit disk?

Thanks.

• $g$ is injective so $g$ preserves boundaries, which means that $g$ sends the open unit disc in one of the components of the complement of the unit circle and $g(a)=0$ means that is the inside, hence the unit disc; the result also follows by maximum modulus since $g$ is holomorphic on the closed unit disc – Conrad Feb 23 at 15:49
• @Conrad this should be an answer. +1 – Fimpellizieri Feb 23 at 15:51
• @Conrad using maximum modulus principle, I get that $\left|g(z) \right| < 1$, so $g(U) \subseteq U$, but how do you get $U \subseteq g(U)$? – Phil Feb 23 at 22:46
• in this case you can easily see it by solving the equation $g(z)=w$ for $|w|<1$ which explicitly computes $g^{-1}$ which is of the same type (ie sends unit circle to itself etc); in general this follows from injectivity which implies that the boundary of $g(U)$ is the full $g(\partial U)$ ie the full circle, which clearly implies that you cannot omit any value as the boundary would be bigger (the technical argument involves Jordan curves, argument principle etc but should be intuitive) – Conrad Feb 23 at 22:58