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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 Riemann 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?

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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.

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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
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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

1 Answer 1

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.

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