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

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.

At $z=1$, the branch value is the complete elliptic integral of the first kindA contour that goes once around $1$ and then to $z$ applies the reflection about $K$, giving

Similarly, the branch value at $-1$ is $-K$, so a loop around $-1$ acts as $R_{-1}(w)=-2K-w$. A contour that loops first around $-1$ and then around $1$ therefore gives

Finally approach the interval $(1,1/k)$ through the upper half-plane. Thereso the branch value at $1/k$ isIt follows that a loop around $1/k$ acts asA contour that loops first around $1$ and then around $1/k$ consequently givesHere $L=K^{\prime }(k)$ is the complementary complete elliptic integral of the first kind. These contours exhibit all three requested values.