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I'm currently stuck on the following problem:

Let $f$ be analytic in the set $\{ z \in \Bbb{C}: 0<|z|<1\}$. If $f$ is real in the unit circle $\{z \in \mathbb{C}: |z|=1\}$, then show that: $$f(z)=\overline{f\left ( \frac{1}{\overline{z}} \right )} \ \forall z \in \Bbb{C}$$

I was given, as a hint, to define $\phi(z)=\frac{z-i}{z+i}$ and consider the study $f\circ \phi$. I observed that: $$\overline{\phi(z)}=\frac{1}{\phi(\overline{z})}$$

Which aesthetically resembles what I have to prove, but I'm unsure on how to proceed.

Any thoughts?

Thanks in advance!

Reveillark
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    If $f$ is real in the unit disk, then $f$ is constant. Is it perhaps that $f$ shall have a continuous extension to the unit circle, and that is real-valued on the unit circle ${ z : \lvert z\rvert = 1}$? – Daniel Fischer Sep 19 '15 at 20:27
  • @DanielFischer Yes, you're absolutely right, I edited accordingly. – Reveillark Sep 19 '15 at 20:29
  • Okay. The exercise is about the reflection principle. The case of a circular boundary is derived from the case for the boundary on the real axis. Note that $\phi$ maps the upper half-plane biholomorphically to the unit disk. So $f\circ \phi$ is defined on the upper half-plane, extends continuously to the real axis, and the extension is real-valued on $\mathbb{R}$ … – Daniel Fischer Sep 19 '15 at 20:33
  • @DanielFischer So by the reflection principle I can extend $f\circ \phi$ to all of $\Bbb{C}$ defining $g: \Bbb{C}\to \Bbb{C}$ by $g(z)=(f\circ\phi) (z)$ if $z \in \Bbb{C}^+$ and $g(z)=\overline{(f\circ \phi)(\overline{z})}$. Sorry, but I'm not seeing how this helps. – Reveillark Sep 19 '15 at 20:46
  • Oh, sorry, I forgot that $f$ is not defined at $0$, so $f\circ \phi$ is not defined at $i$, and $g$ is then defined on $\mathbb{C}\setminus {i,-i}$. Now you have $g$, look at $g\circ \phi^{-1}$. – Daniel Fischer Sep 19 '15 at 20:49
  • @DanielFischer My internet connection died before I could reply, so I had no choice but to think harder on the problem and ended up solving it using the reflection principle + analytic continuation. Thank you very much for your help! – Reveillark Sep 19 '15 at 21:55
  • For a change, that's an internet connection that died at the right time ;) Will you write an answer? – Daniel Fischer Sep 19 '15 at 21:58
  • I hadn't considered it, but I guess I could write an answer for completeness sake. – Reveillark Sep 19 '15 at 22:02

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