2
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This question is in light of a previous question

$f$ be a nonconstant 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.

2 is wrong by Liouville's theorem.

The author has used $f(z)=e^z$, (holomorphic and $f(0)=1$) to show that option 4 is wrong. Isn't $|f(z)|=1$ only for $z=0$ in this case? That means there are only finite (more specifically one) point inside unit disc such that $|f(z)|=1$.

But someone has proved that 1 is the right option.

Where am I going wrong?

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3
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With $f(z) = e^z$, $|f(z)| = e^{\operatorname{Re}(z)}$. Thus $|f(z)| = 1$ if and only if $\operatorname{Re}(z) = 0$, i.e., $z$ is purely imaginary. There are infinitely many imaginary numbers in the unit disk.

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  • $\begingroup$ Thank you. That example alone is enough to show that 1 is the right option then, isn't it?. $\endgroup$ – Jesse P Francis Mar 27 '15 at 15:29
  • 2
    $\begingroup$ For a multiple-choice test, it would be enough by process of elimination. :-) $\endgroup$ – kobe Mar 27 '15 at 15:31

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