# Showing that the image of a function is $\mathbb{C}$ if it satisfies a nice functional equation

Let $f$ be entire and non-constant. Assuming $f$ satisfies the functional equation $f(1-z)=1-f(z)$, can one show that the image of $f$ is $\mathbb{C}$?

The values $f$ takes on the unit disc seems to determine $f$...

Any ideas?

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If $f({\mathbb C})$ misses $w$, then it also misses $1-w$. The only case where $w = 1-w$ is $1/2$, which is $f(1/2)$. Now use Picard's "little" theorem.
Since the function is entire, it misses at most one point by Picard's little theorem. If the point is say, $y$, then $1-y$ must be in the range, unless $y=1-y$. But, in this case, $y=\frac{1}{2}$ and we can deduce that $f(\frac{1}{2})=\frac{1}{2}$. So, $y\neq 1-y$. So, if there is a $z$ such that $f(z)=1-y$, then, $f(1-z)=1-f(z)=y$. So, the range of $f$ is $\mathbb C$.