# Thread: Complex Analysis -- functions on a connected set

1. ## Complex Analysis -- functions on a connected set

Suppose that U is a simply-connected open domain in C and assume
that $\displaystyle f,g : U \rightarrow U$ 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 $\displaystyle f(z_i) = g(z_i)$ for i = 1, 2 and $\displaystyle z_1 \neq z_2$ 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.

2. ## Re: Complex Analysis -- functions on a connected set

Let $\displaystyle H$ be the upper half-space. We know that automorphisms of $\displaystyle H$ are of the form

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

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

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

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

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

Conclude by the Riemann mapping theorem.