# A Conformal Mapping Question

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!

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Are you assuming that $f$ is bijective? (Otherwise, the definite article "the" in "the conformal map" seems uncalled for.) – Charles Staats Apr 27 '11 at 3:22
@Charles: Good point. I had assumed bijective was intended. Otherwise I think $f(z)=-\frac{1}{i-z}-i$, with $U$ some small disk centered at $0$, would be a counterexample. – Jonas Meyer Apr 27 '11 at 3:32
Yes. The "the" is simply meant to emphasize that any map which satisfies the stated properties is unique. – Jonathan Gleason Apr 27 '11 at 3:33
Just to clarify, I meant yes, $f$ is assumed to be bijective. – Jonathan Gleason Apr 27 '11 at 3:34
what is $z^*$ by the way? just $\bar{z}$? – La Belle Noiseuse Apr 28 '13 at 9:18

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.
That does it. I was not familiar with the important fact that if $f$ is holomorphic, so is $f(z^*)^*$. This makes the problem very easy. Thanks so much for the help! – Jonathan Gleason Apr 27 '11 at 3:40
by the is it the way to see that $\bar{f(\bar{z})}$ is analytic if $f(z)$ is analytic? here is how I think: $f(z)=\sum_{n=0}^{\infty}a_n z^n$, so $f(\bar{z})=\sum_{n=0}^{\infty}a_n \bar{z}^n$ and $\bar{f(\bar{z})}=\sum_{n=0}^{\infty}a_n{z}^n$ as $\bar{z}^n=\bar{z^n}$ ? – La Belle Noiseuse Apr 28 '13 at 9:30
@Tsotsi: I believe I have answered about 2% (or 1 in 50) of the questions to date tagged complex-analysis. I first learned complex analysis while taking a class at a university, and the textbook for the course was by J.B. Conway, but there are many other good ones. I don't have much to tell you. Regarding showing that $\overline{f{\overline z}}$ is analytic if $f$ is, there's a question about that here. The second expression you gave for $f(\overline z)$ is incorrect. If $f(z)=\sum a_n z^n$ then $\overline{f(\overline z)}=\sum \overline{a_n}z^n$. – Jonas Meyer Apr 29 '13 at 0:52