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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.)2011-04-27
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    @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.2011-04-27
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    Yes. The "the" is simply meant to emphasize that any map which satisfies the stated properties is unique.2011-04-27
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    Just to clarify, I meant yes, $f$ is assumed to be bijective.2011-04-27
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    what is $z^*$ by the way? just $\bar{z}$?2013-04-28
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    @Tsotsi Yes, the complex conjugate.2013-04-28

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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.

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    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!2011-04-27
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    @GleasSpty: You're welcome.2011-04-27
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    @JonasMeyer I see you The guy answers almost all the complex ana questions, teach me some complex analysis and tell how to be a good complex analysis problem solver :) how did you prepared yourself on complex analysis2013-04-28
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    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 $\bar{z}^n=\bar{z^n}$ ?2013-04-28
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    @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](http://math.stackexchange.com/q/102885). The second expression you gave for $f(\overline z)$ is incorrect. If $f(z)=\sum a_n z^n$ then $\overline{f(\overline z)}=\sum \overline{a_n}z^n$.2013-04-29
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    @JonasMeyer Thank you2013-04-29