There are exactly two holomorphic functions $h:\mathbb{C}\setminus[0,1]$ such that $(h(z))^2=z(z-1)$. Let $\Omega:=\mathbb{C}\setminus [0,1]$. I want to prove that the function $z\mapsto z(z-1)$ has exactly two holomorphic square roots in $\Omega$.
In $\mathbb{C}\setminus (-\infty,1]$, I know that we can use the principal branch of the logarithm to construct
$$h_{\pm}(z)=\pm\exp\left(\frac{1}{2}\text{Log}\: z+\frac{1}{2}\text{Log}\:(z-1)\right).$$
Surely $h_+$ and $h_-$ are square roots of $z\mapsto z(z-1)$ in $\mathbb{C}\setminus (-\infty,1]$. Then I have two questions:

1) How could I extend $h_\pm$ to $\Omega$?
2) How can I be sure that there are no other square roots?

 A: *

*Fix $z_0 \in \Omega$ and choose $a \in \mathbb{C}$ such that $a^2 = z_0 (z_0 - 1)$. Then define $h(z)$ by
$$ h(z) = a \exp \left\{ \frac{1}{2} \int_{z_{0}}^{z} \left( \frac{1}{\xi} + \frac{1}{\xi-1} \right) \, d\xi \right\}, $$
where the path of integration is chosen as any nice curve in $\Omega$ that starts at $z_0$ and ends at $z$. Since we have $\frac{1}{2} \int_{\gamma} \left( \frac{1}{\xi} + \frac{1}{\xi-1} \right) \, d\xi \in 2\pi i \mathbb{Z}$ for any nice closed curve $\gamma \subseteq \Omega$, ambiguity from arbitrary choice of path is cancelled out by exponentiation. So $h(z)$ is well-defined on $\Omega$. Also,
$$ \frac{d}{dz} \frac{z(z-1)}{h(z)^2}
= \frac{1}{h(z)^2} \left( 2z - 1 - 2z(z-1) \cdot \frac{1}{2}\left( \frac{1}{z} + \frac{1}{z-1} \right) \right) = 0 $$
and hence $z(z-1)/h(z)^2$ is constant with the value $z_0(z_0-1)/h(z_0)^2 = 1$. This proves that $h(z)$ is a square root of $z(z-1)$ on $\Omega$.

*Next, let $g(z)$ be defined on a connected domain $\Omega' \subseteq \Omega$ and assume that $g(z)^2 = z(z-1)$ holds. Then $g(z)/h(z) \in \{-1, 1\}$ for each $z \in \Omega'$. Since $g(z)/h(z)$ is continuous and $\Omega'$ is connected, we must have either $g(z)/h(z) \equiv 1$ on $\Omega'$ or $g(z)/h(z) \equiv -1$ on $\Omega'$. Therefore $g(z) = \pm h(z)$.
Summarizing, any square root of $z(z-1)$ on a connected subdomain of $\Omega$ must equal either $h$ or $-h$. In this very sense, there are exactly two square roots of $z(z-1)$ on $\Omega$.
