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.

Possible Duplicate:
Let f(z) be entire function. Show that if $f(z)$ is real when $|z| = 1$, then $f(z)$ must be a constant function using Maximum Modulus theorem

I'm having trouble proving that an analytic function that takes on only real values on the boundary of a circle is a real constant. I started by writing

$f(r, \theta) = u(r, \theta) + i v(r, \theta)$

By definition, $v(r, \theta ) = 0$, so $\frac{d}{d\theta} v = 0$, and in fact, the nth derivative of $v(r,\theta)$ with respect to $\theta$ is 0. The Cauchy Reimann equations in polar coordinates imply that $\frac{d}{ dr} u(r, \theta ) = 0$

Unfortunately, I'm stuck here - I think I need to prove that all nth derivatives of u and v with respect to both r and \theta$ are 0, so that I can move in any direction without inducing a change in f, but at this point I'm stuck. I've played around with this a fair bit but keep running in circles (no pun intended), so there must be something simple that I am missing. What am I doing wrong, and how does one complete the proof?

share|improve this question

marked as duplicate by N. S., Quixotic, Austin Mohr, Davide Giraudo, Phira Dec 11 '12 at 11:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4  
One nice way to see this is to observe that the imaginary part of your function is harmonic, hence it satisfies the maximum principle. You therefore get $v(x,y) \le 0$ for $x^2 + y^2 \le 1$, and by the mean value theorem, $v(0,0) = 0$, hence $v$ must be constant. But then $f$ must be constant too (follows from Cauchy-Riemann equations). –  Mauro Porta Dec 10 '12 at 23:09
1  
    
For contrast, $f(z)=i\dfrac{1+z}{1-z}$ maps the unit circle onto $\mathbb R\cup\{\infty\}$. –  Jonas Meyer Dec 11 '12 at 5:09

2 Answers 2

You know that if $f=u+iv$, then $u$ and $v$ are harmonic. Now, by assumption you have that $v$ is zero on the boundary of the disk. But, by the two extrema principles, you know that the maximum and minimum of $v$ occur on the boundary of your disc, and so clearly this implies that $v$ is identically zero. Thus, $f=u$, and so $f$ maps the disk into $\mathbb{R}$. But, by the open mapping theorem this implies that $f=u=\text{constant}$ else the image of the disk would be an open subset of $\mathbb{C}$ sitting inside $\mathbb{R}$ which is impossible.

share|improve this answer

I see my error - I did use the mean value theorem to conclude that v(0,0) was zero, but did not use the maximum modulus principle to conclude that the real part was necessarily zero everywhere inside the disk and then use the Cauchy-Riemann equations to show that u(r, \theta$ ) was constant on the disk. Instead, I tried to directly manipulate the Cauchy-Riemann equations to show that all derivatives of u and v were 0, so that the function was necessarily 0. I'm less comfortable with the open mapping theorem, but yes, this works as well. Thanks a mill.

share|improve this answer
    
This should be a comment, not an answer. –  Austin Mohr Dec 11 '12 at 3:47
1  
@AustinMohr: not when you don't have enough reputation! –  Martin Argerami Dec 11 '12 at 4:58

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