Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 13E / b / i / Solution

The additional branch points $\pm 1/k$ lie on the two deleted rays because $0<k<1$. On the same simply connected domainchoose both square roots to equal $1$ at zero. Their product has a single-valued holomorphic reciprocal on $D$, so its integral from zero defines the single-valued branch $H$ of the elliptic integral of the first kind.

An unrestricted contour may wind around any of $-1,1,-1/k,1/k$. Continuation around one such point flips one square root, so the value can change by square-root monodromy. Thus $h$ is multivalued.