Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 13E (Further Complex Methods)

PutThis slit complex plane is a simply connected domain. On $D$ there is one branch of $\sqrt{1-z^2}$ that equals $1$ at zero, and its reciprocal is holomorphic. The primitive of a holomorphic function on a simply connected domain therefore makes the integral from zero path-independent, so $G$ is single-valued.

For unrestricted paths, going once around either branch point changes the sign of the square root. The resulting analytic continuation therefore need not return to the original germ, and the value of $g(z)$ can depend on the path.

The endpoint values of the principal branch areBy the monodromy reflection at a square-root branch point, continuation around $-1$ acts on a value $w$ aswhile continuation around $1$ acts as

The contour $C_1$ makes the first continuation and then follows the principal path to $z$, soThe contour $C_2$ continues successively around $-1$ and $1$. Hence

Every path can change sheets only by applying the two reflectionsTheir products satisfyThus an even number of reflections gives a translation by an integral multiple of $2\pi$, while an odd number gives a translated reflection. All possible values are thereforein agreement with the branches of the complex inverse sine.

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.

Part (b) gives paths with the same endpoint $z$ but integral valuesApplying the local inverse and then continuing it shows thatSince $K,L>0$, the first period is real and the second is purely imaginary, so they are linearly independent over $\mathbb{R}$. Hence $\mathcal{H}$ is an elliptic function with two periodsIndeed, it is the Jacobi elliptic sine, and these generators give its period lattice.

The inverse is nonconstant; locally at zero,If the assumed meromorphic function $\mathcal{H}$ had no poles, it would be entire. It is bounded on the compact closure of a fundamental parallelogram, and its two periods then make it bounded throughout the complex plane. The Liouville theorem would force it to be constant, a contradiction. Therefore $\mathcal{H}$ has at least one pole, as stated by the general result that a nonconstant elliptic function has a pole.