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$f$ be a non-constant holomorphic in unit disk such that $f(0)=1$. Then it is necessary that

  1. there are infinitely many points inside unit disk such that $|f(z)|=1$

  2. $f$ is bounded.

  3. there are at most finitely many points inside unit disk such that $|f(z)|=1$

  4. $f$ is rational function.

Counter examples for $2,4$ are ${1\over z-1},e^z$

If there are infinitely many points in the open disk such that $|f(z)|=1$ then that set is being bounded. Since there are infinite points so it must have a limit point in it and $f$ will be bounded. And hence by Liouvilles Theorem, it is constant. This is a contradiction

So $3$ is the correct statement. Thank you for confirming.

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  • $\begingroup$ Note that the question is about $\lvert f(z)\rvert = 1$, not about $f(z) = 1$. $\endgroup$ Jan 6 '14 at 14:57
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Hint: For $1,3$ use the Open Mapping Theorem with the open unit disk to prove that there are infinitely many points.

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  • $\begingroup$ i dont understand,explain $\endgroup$
    – Marso
    Jan 6 '14 at 15:31
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    $\begingroup$ Maybe I misunderstood your question. I wanted to give an argument why for a non-constant holomorphic function $f$ on the unit disc $\Bbb E$ with $f(0)=1$ there are infinitely many $x \in \Bbb E$ with $|f(x)|=1$, that is $f(x)$ is on the unit circle: As $f$ is non-constant, by the Open Mapping Theorem the unit disc is mapped to an open set $f(\Bbb E)$ in $\Bbb C$ with $1 = f(0) \in f(\Bbb E)$. Therefore, it contains an open neighborhood of $1$. Intersect this neighborhood with the unit circle to get infinitely many points in the image with $|f(x)| = 1$. $\endgroup$
    – benh
    Jan 6 '14 at 15:56
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Liouville's theorem can be used to contradict 2.

2, 3 and 4 are wrong, if we consider the example $e^z$ mentioned in the question.

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