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Let $f:\mathbb{C} \to \mathbb{C}$ be a non constant entire function such that $f(1-z)+f(z)=1$ for all $z\in \mathbb{C}$. Then prove that $f$ is surjective.

It can be solved trivially by Picard's theorem but I don't want to use it. Is there any elementary way in which we can solve it ?

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    $\begingroup$ Interesting observation: if $g(z) = 1 - z$, then this states that $f \circ g = g \circ f$, and that $f = g \circ f \circ g$ (since $(g \circ g)(x) = x$). I don't know if this is useful, though. $\endgroup$ – Omnomnomnom Jun 28 '16 at 3:12
  • $\begingroup$ non constant entire function may not onto for example $e^{z}$. How picard theorem helps you? $\endgroup$ – neelkanth Jun 28 '16 at 5:22
  • $\begingroup$ Why do you think $e^{1-z}+e^{z}=1$ ? read the question properly please :) $\endgroup$ – dragoboy Jun 28 '16 at 5:28
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    $\begingroup$ If you really wanna know how picard's help here then let me tell you: By Picard's theorem we know $f$ can't give at most one value. Suppose $f$ can't give $a$, so it can't give $1-a$ from the given condition. And so by picard's theorem we must have $a=1-a \implies a=\frac{1}{2}$ but $f(\frac{1}{2})=\frac{1}{2}$, contradiction ! $\endgroup$ – dragoboy Jun 28 '16 at 5:31

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