TITLE: Showing that the image of a function is $\mathbb{C}$ if it satisfies a nice functional equation
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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 $\mathbb C$.