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$f$ is analytic in $\mathbb{D}$ and continuous on $\mathbb{D}$ closure. If $f(e^{i\theta})$ is a real number for $\theta$ in between $0$ to $2\pi$. Prove that $f$ is constant. Also I want to know what will happen if $\theta$ is in between $0$ to $\pi$?

I don't exactly know where to start. Please help.

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Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Also, many would consider your post rude because it is a command ("Prove..."), not a request for help, so please consider rewriting it. – Zev Chonoles Dec 19 '12 at 17:50
@ZevChonoles, sorry I do not mean to be rude. Hope you like this better. I just started learning from this website. Thanks for the suggestion. – Deepak Dec 19 '12 at 18:12
@Deepak: Don't worry about it - there are always some things to get used to in a new community. Now you know for any future questions! :) – Zev Chonoles Dec 19 '12 at 18:13
@ZevChonoles Absolutely. – Deepak Dec 19 '12 at 18:18
up vote 0 down vote accepted

One route is to use the Poisson kernel representation of $f$. This can be derived in may ways, eg, Derivation of the Poisson Kernel from the Cauchy Formula. (This formulation works for functions analytic on $B(0,R)$, but can be easily extended to deal with functions analytic on $B(0,R)$ and continuous on $\overline{B}(0,R)$.)

Let $B$ be the open unit ball. For $0\leq r < 1$, the Poisson kernel representation gives $f(r e^{i\theta}) = \frac{1}{2 \pi} \int_{-\pi}^\pi P_r(\theta -t) f(e^{it}) dt$, where $P_r(\theta) = \text{Re} \frac{1+re^{i \theta}}{1-re^{i \theta}}$. It follows from this that if $f(e^{i t})$ is real for all $t$, then $f(r e^{i\theta})$ is real for all $0\leq r < 1$. Hence $f(z)$ is real for all $z \in B$.

If $f(z)$ is real for all $z \in B$, it follows that $f'(z) = 0$ for $z \in B$ (if $f'(z_0) \neq 0$, then by looking at $\phi(t) = f(z_0+i t \overline{f'(z_0)})-f(z_0)$, it is straightforward to contradict $f(z)$ being real). Since $B$ is connected, it follows that $f$ is constant on $B$.

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Hint: Since $f(e^{i\theta})\in \mathbb{R}$ then $f(z)\in \mathbb{R}$ for $z\in D$ (why?) . If $f=u+iv$ then $v\equiv 0$ and...

Proof of the first part. Remember that $z\in \partial D\iff z=e^{i\theta}$ for some $\theta\in [0,2\pi)$. Then $f(z)\in \mathbb{R}$ for $z\in \partial D$. Use the identity theorem...

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At@Nameless Nameless. I still confuse. Can you prove more rigorously please? – Deepak Dec 19 '12 at 18:19
Which part do you want me to prove more rigorously? – Nameless Dec 19 '12 at 18:20
First part please. – Deepak Dec 19 '12 at 18:21
Try the Poisson kernel. – copper.hat Dec 19 '12 at 18:22
@Nameless: Are you using $D$ to be the unit disk or its boundary? – copper.hat Dec 19 '12 at 18:24

Extend $f$ to $|z| > 1$ by $f(z) = \overline{f(1/\overline{z})}$. Note that the extended function is analytic in $\{z: |z| > 1\}$ as well as $\{z: |z| < 1\}$ and continuous on all of $\mathbb C$. By Morera's theorem it is an entire function. Now use Liouville's theorem.

EDIT: for a nice example where $f(e^{i\theta})$ is real for $0 \le \theta \le \pi$, take $$f(z) = \left( \sqrt {{\frac {i-iz}{z+1}}}+1 \right) ^{-1}$$ (using the principal branch of the square root) with $f(-1) = 0$.

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