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How can I prove this. I could not use $\Im(w)<0$ condition in Liouville's theorem.

Let $f(z)$ be an entire function and assuming that $f(z)$ does not take values in $\Im(w)<0$ show that $f$ is identically zero.

Thanks.

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  • 3
    $\begingroup$ You have to assume non-constant. $\endgroup$ – Davide Giraudo Jun 4 '12 at 16:42
  • 2
    $\begingroup$ Consider $g\circ f$ where $g$ is a suitable fractional linear transformation. $\endgroup$ – user31373 Jun 4 '12 at 16:45
  • $\begingroup$ What can you say about $g(z) = \frac{f(z)}{f(z)+i}$. $\endgroup$ – copper.hat Jun 4 '12 at 16:51
  • $\begingroup$ How g can be used in thm ? $\endgroup$ – Brhn Jun 4 '12 at 17:00
  • $\begingroup$ $g$ is bounded and entire. What can you say about such functions? And what does that say about $f$? $\endgroup$ – copper.hat Jun 4 '12 at 17:06
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Let $g(z)=e^{if(z)}$. Then $g$ is entire and $$|g(z)|=|\exp((\Re f(z)+i\Im f(z))i)| =|\exp\left(i\Re f(z)-\Im f(z)\right)|=e^{-\Im f(z)}\leq 1.$$ By Liouville theorem, $g$ is constant hence $e^{if(z)}=C$ and $f'(z)e^{if(z)}=0$ so $f$ is constant (but not necessarily $0$).

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Suppose $f$ is non constant. Then $\mathcal{Im}(f)$ is open so we can assume $\mathcal{Im}(f) \subset I(w) > 0$

$ \varphi :z \mapsto \frac{z - i }{z + i}$ is a bijection from $\mathbb{D}$ to $I(w) > 0$

So $\varphi^{-1} \circ f $ is an entire function which is bounded so by Liouville it is constant and then $f$ is constant.

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