# Complex Analysis -- functions on a connected set

• Sep 15th 2011, 11:54 AM
harbottle
Complex Analysis -- functions on a connected set
Suppose that U is a simply-connected open domain in C and assume
that $f,g : U \rightarrow U$ are one-to-one and onto maps which are holomorphic mappings with the property that f' and g' are non-zero for all points of
U. Prove that if $f(z_i) = g(z_i)$ for i = 1, 2 and $z_1 \neq z_2$ then f = g.

I am not sure where to start on this one. I know that if f and g are equal on a set which is not discrete then they are the same, but in this case the set is just two elements. I'm not sure what piece of the puzzle I am missing -- any help would be appreciated.
• Sep 15th 2011, 03:27 PM
Jose27
Re: Complex Analysis -- functions on a connected set
Let $H$ be the upper half-space. We know that automorphisms of $H$ are of the form

$f(z)=\frac{az+b}{cz+d} \qquad a,b,c,d\in \mathbb{R} \qquad ad-bc=1$

then (working with $f\circ g^{-1}=h$) assume $h$ is such a function with two fixed points. Without loss of generality assume $h(i)=i$ then $h(z)=\frac{az+b}{-bz+a}$. Let $w\in H$ be the other fixed point then

$w=h(w)=\frac{aw+b}{-bw+a}$

and this leads to, assuming $b\neq 0$, $w=i$ or $w=-i$ so we have $b=0$ or $h(z)=z$.

Now, since automorphisms of $\mathbb{C}$ are affine transformations the result in this case is obvious.

Conclude by the Riemann mapping theorem.