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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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    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).2012-12-10
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    http://math.stackexchange.com/questions/226786/let-fz-be-entire-function-show-that-if-fz-is-real-when-z-1-then-f2012-12-11
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    For contrast, $f(z)=i\dfrac{1+z}{1-z}$ maps the unit circle onto $\mathbb R\cup\{\infty\}$.2012-12-11

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