I have a holomorphic function $f,g: \mathbb C \to \mathbb C$ such that $e^{f(z)} + e^{g(z)} = 1$ for any $z$ in $\mathbb C$. Does anyone have any tips that to show that this is bounded? Or check whether it is bounded?
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1Show what is bounded? Surely not $f$ or $g$, since you said they are both holomorphic on $\mathbb C$? – MPW Apr 11 '17 at 19:28
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(cont'd) Unless they are constant – MPW Apr 11 '17 at 19:42
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@MPW, I assume he means to apply Liouville and show such functions are constant. – Apr 11 '17 at 19:56
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Use Picard's theorem. If an entire function omits two points it is constant.
Firstly $e^{g(z)}$ and $e^{f(z)}$ can't equal zero, because $e^z$ never equals zero. But just as well, $e^{f(z)} = 1 - e^{g(z)}$ so $e^{g(z)}$ can't equal $1$. Therefore $e^{g(z)}$ is an entire function that omits two points, it must be constant by Picard. Therefore $e^{f(z)}$ is constant. Both are obviously bounded.
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Hello James, thanks for the reply! This is very similar to an assignment question I was assigned but that question specifically told me that I could omit one point for each e^f(z) and e^g(z). I was never taught Picard's thoerem. Thanks for the hints anyways!! – Wilson Apr 12 '17 at 03:57