TITLE: analytic functions from square to unit disk
QUESTION [5 upvotes]: Let $f$ be an analytic function from $\{z; -1 < \Re(z) < 1, -1 < \Im(z) < 1\}$ to $\{z; |z| < 1\}$. If $f(0)=0$ and $f$ is one-one and onto, should $f(i\ z)=i\ f(z)$ for each $z$? I tried to show that $f(i\ z)-i\ f(z)$ is a constant, but it seems that I could not use  Liouville Theorem. 
Thank you very much.

REPLY [6 votes]: Assuming you mean the open square, then yes.
Let $g(z)=f(iz)$ and $h(z)=if(z)$.  Then $g$ and $h$ are analytic bijections from the square to the disk such that $g(0)=h(0)=0$ and $g'(0)=h'(0)=if'(0)$.  This implies that $k=g\circ h^{-1}$ is an analytic bijection of the disk such that $k(0)=0$ and $k'(0)=1$.  By Schwarz's lemma, $k(z)=z$ for all $z$ in the disk, and therefore $g=h$.