1

Suppose $f'$ is holomorphic in $\overline{\mathbb D}$, $\mathbb D=\{z\mid |z|<1\}$ and $|f(z)|=1$, Prove $f$ is a rational function.

As you can see, I didn't get it. Why does it say $f'$ is holomorphic not $f$ is holomorphic? Also, I have no idea how to prove a holomorphic function is a rational function, in what way? Please give me some clues, thanks.

John Smtih
  • 206
  • 1
    $;f;$ is holomorphic automatic where $;f';$ exists... – DonAntonio Jun 07 '16 at 07:24
  • Don't your conditions imply that $f$ is constant? –  Jun 07 '16 at 07:27
  • http://math.stackexchange.com/questions/147834/why-a-holomorphic-function-with-constant-magnitude-must-be-constant –  Jun 07 '16 at 07:28
  • Where is the different to your previous question http://math.stackexchange.com/questions/1815795/suppose-f-is-holomorphic-in-overline-mathbb-d ? – Martin R Jun 07 '16 at 08:24

1 Answers1

1

I assume you mean $|f(z)|=1$ just on $\partial\mathbb D$. Extend $f$ to all of the extended complex plane using $$\tilde f(z)=\left\{\begin{array}{ll}f(z),&z\in\mathbb D\\1/f(1/z),&z\notin\mathbb D\end{array}\right.$$ This is clearly holomorphic inside $\mathbb D$, and meromorphic ouside of $\mathbb D$. You can argue it is meromorphic on the whole space using Morera's theorem (since a closed curve intersecting the boundary you can break up into two curves in the two regions). Then use the theorem that meromorphic functions in the extended complex plane are rational to conclude that $\tilde f|_\mathbb D=f$ is rational. If you haven't seen this theorem, you can prove it by essentially subtracting off the poles (of which there must be finite) until you get a bounded entire function.

Funktorality
  • 3,221