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For $D$ the open unit disk in the complex plane and $f: D \rightarrow \Bbb{C}$ holomorphic and $1-1$, does there exist a holomorphic function $g: D\rightarrow\Bbb{C}$ such that $d^2=f'$ ?

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    $\begingroup$ Do you mean $g^2 = f$? $\endgroup$ – Christopher A. Wong Feb 12 '13 at 18:34
  • $\begingroup$ Nope, $f'$ is correct. $\endgroup$ – Mike Feb 12 '13 at 19:11
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The answer is yes.

1) as $f$ is holomorphic and bijective, it is biholomorphic ;

2) if $f$ is biholomorphic, then $f'$ never vanishes ;

3) you can show that $f'(\mathrm D)$ is simply connected. As it is also not containing zero, there exists a well-defined square-root function on $f'(\mathrm D)$.

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  • $\begingroup$ Thank you! This seems to be what I was looking for, I didn't know about biholomorphic functions. $\endgroup$ – Mike Feb 12 '13 at 19:15
  • $\begingroup$ I was writing it out properly and I just noticed: $f$ is not bijective, it's only injective, so it's not biholomorphic. $\endgroup$ – Mike Feb 13 '13 at 20:29
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    $\begingroup$ We don't care, injective implies bijective on the image. $\endgroup$ – Damien L Feb 13 '13 at 20:31
  • $\begingroup$ Oh right, we're talking about $f(D)$ not $\Bbb{C}$. I think I got it the first time I read it but when I did today I got confused, sorry. $\endgroup$ – Mike Feb 13 '13 at 20:39

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