Holomorphic square root ($g^2=f$)

A holomorphic square root of a nonvanishing holomorphic function $f$ is a holomorphic function $g$ satisfying $g^2=f$. It exists locally. On a connected open domain it exists globally exactly when the winding number of $f\circ \gamma$ around zero is even for every closed path $\gamma$ in the domain. Continuing a local square root once along such a path multiplies it by $(-1{)}^m$, where $m$ is that winding number. Thus the obstruction concerns parity, whereas a holomorphic logarithm needs every such winding number to be zero.

On $\mathbb{C}\setminus \{0\}$, the function $f(z)=z$ has no global holomorphic square root: its value winds once on the unit circle, and the lift shown under monodromy changes sign. In contrast, $f(z)=z^2$ winds twice there and has global square roots $g(z)=z$ and $g(z)=-z$. It still has no holomorphic logarithm on that domain. This distinguishes the two global continuation conditions.

The branch points are $\pm 1$; infinity has two unbranched meromorphic choices. On the complex plane slit along $[-1,1]$, normalization $f(z)\sim z$ at infinity selects a holomorphic square root with boundary values $f(x+i0)=i\sqrt{1-x^2}$ and $f(x-i0)=-i\sqrt{1-x^2}$ for $-1<x<1$. On the exterior real intervals, $f(x)=\mathrm{sgn}(x)\sqrt{x^2-1}$. A counterclockwise contour integral around the finite branch cut satisfies $\oint d\zeta /f(\zeta )=2\pi i$: the upper lip runs right to left, the lower lip left to right, and the endpoint arcs contribute $O(\sqrt{\epsilon })$. The opposite global branch reverses these signs.

An even-degree polynomial with all roots in $|z|<R$ admits a holomorphic square root throughout $|z|>R$. Factor out its even power of $z$ and use the convergent logarithm of each $1-a_j/z$ to construct it. Even degree removes the annular winding obstruction. The two choices differ by a global sign.