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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.

  • 1
    $E$ is simply connected.2011-12-08
  • 6
    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.2011-12-08
  • 0
    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 $E$.2011-12-08
  • 0
    I don't understand how the fact that E is simply connected help me.2011-12-08
  • 1
    Then have another look at your favorite complex analysis textbook. Which one is it, by the way?2011-12-08
  • 1
    @Andrea, so is $\mathbb C$ itself, but there's no solution there (assuming that $f^2(x)$ means $f(x)^2$).2011-12-08
  • 0
    First you say $f$ is analytic in $E$, but then you apply $f$ to every $z \in \mathbb C$ ... what did you really mean?2012-01-19

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