Function $ f $ defined on unit disk $ D $ has real values on $ \partial D $, why is $ f $ real valued? I have a function $ f $ that is analytic on the unit disc and real valued on the unit circle. I want to prove that $ f $ is real valued on the unit disk. 
I know that I need to use the identity theorem, but I am unsure how. The only thing I can think to do is:
Let $ g $ be some real valued function s.t. $ g(e^{i\theta}) = f(e^{i\theta}) $ then $  f = g $, but this argument obviously isn't sound. Any help would be greatly appreciated. 
 A: Note: this answers a different question (assuming $f$ is holomorphic).  I leave it here for now because it may have some value anyway.
I assume that $f$ is continuous on the closed disc.  One approach uses Schwarz' reflection principle: The function $f$ can be extended to an entire function by defining 
$$ f(z) = \overline{f(1/\overline{z})} $$
for $|z| > 1$.  (The hardish part of the reflection principle is that the resulting function is indeed holomorphic also for points on the unit circle.)  Now $f$ is extended to a bounded entire function (since it is bounded on the compact closed unit disc) so it must be constant.
Another approach uses the open mapping theorem.  Suppose $f$ is not constant.  Then the image of the open unit disc is open and bounded (since $f$ is continuous on the closed unit disc).  In particular it must have a non-real boundary point.  This boundary point must be attained by $f$ on the compact closed unit disc.  But $f$ is an open mapping on the open unit disc so this non-real boundary point must be attained on the unit circle.  Contradiction.
A: By analytic on the unit disk, I assume you mean holomorphic on a neighborhood of the closed unit disk. If it's just analytic on the open unit disk, the statement is obviously false, as $f(z)=z $ for $|z|<1$ and $f(z)=0$ for $|z|=1$ is a counterexample.
Recall that the real and imaginary parts of a holomorphic function are harmonic.
The function $\mbox{Im} f$ is harmonic on the unit disk and zero on the unit circle. By the maximum/minimum principle for harmonic functions, it follows that $\mbox{Im} f=0$ on the whole unit disk.
