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This question already has an answer here:

Let $f$ be entire and suppose there is a constant $M>0$ such that $|f(z)|>M$ for all $z \in \mathbb C$. Prove that $f$ is constant.

I think this has something to do with Liouville's theorem but not sure how to go about it!

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marked as duplicate by Martin R, amd, zz20s, user228113, colormegone May 6 '16 at 22:30

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    $\begingroup$ What can you say about $\frac{1}{f}$? $\endgroup$ – carmichael561 May 6 '16 at 19:52
  • $\begingroup$ That 1/f > 1/M ? $\endgroup$ – maths.gal May 6 '16 at 19:55
  • $\begingroup$ $1/|f|<1/M$, right? $\endgroup$ – carmichael561 May 6 '16 at 19:55
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Show that $g(z):=\frac1{f(z)}$ is constant.

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Hint : Apply Liouville theorem to $\frac{1}{f}$.

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  • $\begingroup$ How do I know to do that though? $\endgroup$ – maths.gal May 6 '16 at 19:55
  • $\begingroup$ well to apply Liouville theorem your function needs to be entire and it must be bounded. The intuition is that if you have $|f|>M$ then $|\frac{1}{f}|<1/M$. But after yoi have to verify all the other conditions, since $f \neq 0$, $\frac{1}{f}$ is also entire. So the intuitions was true :). $\endgroup$ – Jennifer May 6 '16 at 19:59

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