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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.2011-03-23
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    @lhf, thank you very much. I mean open square.2011-03-23
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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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    @Jonas, thank you very much.2011-03-23
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    @Jonas, but $k(0)=g(0)/h(0)$ where h(0)=0. Why $k(0)=0$?2011-03-24
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    @Jianrong: $h^{-1}$ in this context is the inverse function of $h$, not the reciprocal of $h$.2011-03-24
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    @lhf, thank you very much.2011-03-24
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    If Schwarz's lemma is satisfied, then $k(z)=az$ for some $|a|=1$. How could we show that $a=1$?2011-03-24
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    Jianrong: $k'(0) = a$, specifically.2011-03-24
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    @Steven, thank you very much.2011-03-24