Holomorphic function such that $g(z)^2=z^2-1$ I have the following subset of $\mathbb{C}: $
$A = \left\{ z \in \mathbb{C}: Im(z) > 0 \right\}$
And I have to decide if there exist an holomorphic map $g:A \rightarrow \mathbb{C}$ such that:
$g(z)^2 = z^2-1$
I've thought about using the following theorem:
$g(z)^n = h(z) \iff \frac{1}{n}\int_{\gamma}\frac{h'(z)}{h(z)} \in 2\pi ik$
with $k \in \mathbb{Z}$ and $\forall\gamma$ loop in $A$.
So I try to compute the following integral:
$\int_{\gamma}\frac{2z}{z^2-1} = \int_{\gamma}\frac{2z}{(z-1)(z+1)}$
At this point I try to use the Cauchy Integral Formula but neither $z_1=1$ nor $z_2=-1$ are inside $A$.
And I get stuck.
Anyone can help me? Thank you in advance.
 A: If $\Omega$ is a simply-connected region and $f$ is holomorphic on $\Omega$ with no zeroes, $f$ admits a holomorphic square root. If you have access to this theorem it gives the result. One can also prove this in the case you're working in with elementary results, as below. An alternative approach is to take local square roots and use monodromy.
An early result in complex analysis is that if $\Omega$ is a convex region and $h$ is holomorphic on $\Omega$, then $h$ admits an antiderivative on $\Omega$ - there's a holomorphic function $f$ on $\Omega$ so $f' = h$. Now, as $A$ is convex and $f(z) = z^2 -1$ has no zero on $A,$ let $g$ be holomorphic on $A$ so that $g'(z) = f'(z)/f(z)$ for all $z \in A$, and adjust $g$ by a constant so that $\exp(g(i)) = f(i)$. For any $z \in A$, $$ \frac{d}{dz} f(z)e^{-g(z)} = f'(z) e^{-g(z)} -f(z)e^{-g}\frac{f'(z)}{f(z)} = 0.$$ As $f$ and $e^{g}$ agree at a point and $A$ is connected we conclude that $f(z) = e^{g(z)}$ on all of $A$. A holomorphic square root of $f(z)$ is then given by $e^{g(z)/2}$.
