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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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    @lhf, thank you. Now it is fixed.2011-03-24

1 Answers 1

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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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    @Steven, thank you very much.2011-03-24