3
$\begingroup$

Let $U$ be an open, simply connected subset of $\mathbb{C}$ that contains $0$ and is symmetric about the real axis. Let $f:U\rightarrow D$, where $D$ is the unit disk, be the conformal map such that $f(0)=0$ and f'(0)>0. Is it necessarily the case that $f(z^*)=f(z)^*$?

My guess is that it is true. It seems intuitive and the couple examples I've written down concretely work.

I've been working on this for about an hour and a half now, and the best I've been able to do is reduce it to proving that $f(x)$ is real if $x\in \mathbb{R}$ (the Schwarz Reflection Principle finishes it off).

Any suggestions/hints/pointers/solutions would be greatly appreciated!

  • 0
    @Tsotsi Yes, the complex co$n$jugate.2013-04-28

1 Answers 1

2

Let $g(z)=f(z^*)^*$. Then $g$ is a conformal map from $U$ to $D$ such that $g(0)=0$ and g'(0)=\lim_{h\to0}\frac{f(h^*)^*}{h}=\left(\lim_{h\to0}\frac{f(h^*)}{h^*}\right)^*=f'(0)^*=f'(0). Your use of the definite article in "the conformal map" indicates to me that you can probably take it from there.

  • 0
    @JonasMeyer Thank you2013-04-29