TITLE: Define square root function over the complex numbers.
QUESTION [7 upvotes]: Define a holomorphic function $f\colon\mathbb{C}\setminus[-1,1] \longrightarrow \mathbb{C}$ such that $\forall z \in \mathbb{C}\setminus[-1,1] \ \left( (f(z))^{2} = z^{2} - 1\right)$ and $f(2)=\sqrt{3}$.

REPLY [5 votes]: For any path between $2$ and $z$ that doesn't intersect the real line between $+1$ and $-1$, define
$$
f(z)=\frac{\log(3)}{2}+\int_2^z\frac{\zeta\,\mathrm{d}\zeta}{\zeta^2-1}
$$
Since $f'(z)=\frac{z}{z^2-1}=\frac{1/2}{z-1}+\frac{1/2}{z+1}$, after accounting for the constant of integration at $z=2$, we get that $f(z)=\frac12\log(z^2-1)$ .
Since the sum of the residues of $\frac{\zeta}{\zeta^2-1}=\frac{1/2}{\zeta-1}+\frac{1/2}{\zeta+1}$ at $\zeta=+1$ and $\zeta=-1$ is $1$, the difference of the integral between two different paths that don't intersect the real line between $+1$ and $-1$ must be an integral multiple of $2\pi i$. Thus, $e^f$ is the same over both paths.
Therefore, $e^{f(z)}=\sqrt{z^2-1}$ is well-defined independent of the path taken.