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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!

Jonathan Gleason
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1 Answers1

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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.

Jonas Meyer
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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! – Jonathan Gleason Apr 27 '11 at 03:40
  • @GleasSpty: You're welcome. – Jonas Meyer Apr 27 '11 at 11:46
  • @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 analysis – Myshkin Apr 28 '13 at 09:23
  • 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 $\bar{z}^n=\bar{z^n}$ ? – Myshkin Apr 28 '13 at 09:30
  • @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 00:52
  • @JonasMeyer Thank you – Myshkin Apr 29 '13 at 05:40