Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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 $\bar{z}$? –  Bunuelian Trick Apr 28 '13 at 9:18
show 1 more comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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
    
@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 –  Bunuelian Trick Apr 28 '13 at 9:23
    
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 $\bar{z}^n=\bar{z^n}$ ? –  Bunuelian Trick 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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.