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I am trying to solve this problem:

Does there exist a function $f(z)$, that is analytic at $E=\{x+iy :x>y\}$ and provides $f^2(z)=z$ for every $z \in \mathbb C$.

I have seen a solution that assumes $f(z)=\sqrt z$, but i have a bad feeling about this way.

Do you have a good solution (or convince me that the square root solution is good).

Thanks.

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$E$ is simply connected. –  Andrea Dec 8 '11 at 14:43
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What do you mean by $f^2(z)$? $f(f(z))$ or $f(z) \cdot f(z)$? From context I assume the latter, though I would usually interpret that notation as the former. –  user7530 Dec 8 '11 at 14:55
    
Clearly $f(z)=\sigma(z)\sqrt{z}$ at every point in $E$, where $\sigma:E\rightarrow\{-1,1\}$. You just need to show that you can choose $\sigma$ so the branch cut avoids $E$. –  mjqxxxx Dec 8 '11 at 14:58
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Then have another look at your favorite complex analysis textbook. Which one is it, by the way? –  Did Dec 8 '11 at 23:28
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@Andrea, so is $\mathbb C$ itself, but there's no solution there (assuming that $f^2(x)$ means $f(x)^2$). –  Henning Makholm Dec 8 '11 at 23:45

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