# Existence of a holomorphic function such that…

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 $d^2=f'$ ?

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

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 $f'(\mathrm D)$.

• Thank you! This seems to be what I was looking for, I didn't know about biholomorphic functions. – Mike Feb 12 '13 at 19:15
• I was writing it out properly and I just noticed: $f$ is not bijective, it's only injective, so it's not biholomorphic. – Mike Feb 13 '13 at 20:29
• We don't care, injective implies bijective on the image. – Damien L Feb 13 '13 at 20:31
• 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. – Mike Feb 13 '13 at 20:39