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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.

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Do you mean the open square or the closed disk? As it stands, the answer to your question is no simply because the closed square is compact but the open disk is not. – lhf Mar 23 '11 at 21:34
@lhf, thank you very much. I mean open square. – user Mar 23 '11 at 23:51
@lhf, thank you. Now it is fixed. – user Mar 24 '11 at 1:01
up vote 6 down vote accepted

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 $g=h$.

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@Jonas, thank you very much. – user Mar 23 '11 at 23:51
@Jonas, but $k(0)=g(0)/h(0)$ where h(0)=0. Why $k(0)=0$? – user Mar 24 '11 at 0:58
@Jianrong: $h^{-1}$ in this context is the inverse function of $h$, not the reciprocal of $h$. – lhf Mar 24 '11 at 1:09
@lhf, thank you very much. – user Mar 24 '11 at 1:18
If Schwarz's lemma is satisfied, then $k(z)=az$ for some $|a|=1$. How could we show that $a=1$? – user Mar 24 '11 at 1:30

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