Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S1

The simple poles occur at $z=n-\alpha$, $n\in \mathbb{Z}$, since the derivative of the sine there is $\pi (-1{)}^n$. ThusThere is also a double pole at zero, not included in that list because $0<\alpha <1$. Its residue, needed for the next contour calculation, is the derivative of $\csc \pi (z+\alpha )$ at zero, namely $-\pi \csc (\pi \alpha )\cot (\pi \alpha )$.

On the square, $|z|\ge N$. The identity $|\sin \pi (x+\alpha +iy){|}^2={\sin }^2\pi (x+\alpha )+{\sinh }^2\pi y$ bounds the sine away from zero on the vertical sides because $x=\pm N$; on horizontal sides it grows exponentially. Therefore the whole contour integral is $O(N^{-1})$ or smaller and vanishes. The residue theorem then giveswhich proves $\boxed{\sum _{n\in \mathbb{Z}}(-1{)}^n/(n-\alpha {)}^2={\pi }^2\csc \pi \alpha \cot \pi \alpha }$. Absolute convergence justifies the limiting residue sum.

Separate the $n=0$ term and pair $n$ with $-n$. For small $\alpha$,Compare the constant and quadratic coefficients with the supplied expansion after subtracting ${\alpha }^{-2}$. It follows thatThe paired expansion is uniformly summable on any sufficiently small closed interval of $\alpha$, so coefficient comparison is legitimate.

A conformal map preserves local oriented angles and has a nonzero local scale factor. For an orientation-preserving complex map this is equivalent to holomorphy with $f^{\prime }(z)\ne 0$ at the point concerned. Infinitesimal figures are rotated and uniformly scaled there, but finite shapes need not be preserved: derivative magnitude and direction can vary across a region. Antiholomorphic maps preserve unoriented angles while reversing orientation; the convention being used must be stated.

For real $x\ne -1$, $w=1-2/(x+1)$ is real. Its finite image is the entire real axis except $w=1$, which is approached only as $x\to \pm \infty$. The point $x=-1$ maps to infinity. On the extended complex plane the extended real axis maps onto itself, with $z=\infty$ mapping to one.

The identity $w-1=-2/(z+1)$ gives $\boxed{|w-1|=2/r}$ on $|z+1|=r$. Thus the image circle has center one and radius $2/r$. Points strictly inside the original circle satisfy $|w-1|>2/r$; the center itself maps to infinity. The interior therefore becomes the exterior together with infinity on the Riemann sphere, rather than the interior of the image circle. This inversion is a useful warning against assuming conformality preserves global inside/outside assignments.

For $z=e^{i\theta }$, $w=i\tan (\theta /2)$, which is purely imaginary; $z=-1$ supplies the point at infinity. Conversely, the inverse map is $z=(1+w)/(1-w)$. Requiring $|w|=1$ is equivalent to $|z-1|=|z+1|$, hence $\boxed{\mathrm{Re}z=0}$, including the appropriate infinite point in the extended-plane formulation. The unit circle and imaginary axis are exchanged, but the real axis is not.

Set $z=e^{i\theta }$. The integral becomesThe roots are $z=(-3\pm 2)/(2-i)$, only $z_{+}=-1/(2-i)$ lying inside. The derivative of the quadratic at $z_{+}$ is two, so the residue is $1/(2i)$. Consequently $\boxed{{\int }_0^{2\pi }d\theta /(3+2\cos \theta +\sin \theta )=\pi }$. The denominator is positive since $3>\sqrt{5}$, so no principal-value qualification is required.

Cauchy's theorem says that a holomorphic function integrates to zero around a closed contour homologous to zero in its domain. Thus deforming one contour into another without crossing a singularity preserves the integral. More generally for a meromorphic $f$, the exact condition isEqual winding numbers about every singularity are sufficient. They are not necessary for one particular function because residues can cancel; even arbitrary contours have equal integrals for $f=0$. A topological deformation criterion must not be confused with a necessary condition for accidental equality.

The finite singularities are square-root branch points at $z=\pm 1$. Choose the cut between them and the branch $f(z)\sim z$ at infinity. The two banks then have values $+i\sqrt{1-x^2}$ above and $-i\sqrt{1-x^2}$ below. A counterclockwise loop around the cut traverses its upper bank right-to-left, so its integral is $-2iI$, where $I$ is the required real integral. At large radius,Therefore $\boxed{I=\pi /2}$. Infinity is not another square-root branch point: this branch has an ordinary pole there on its sheet. Small detours about the finite endpoints vanish as their radius to power $3/2$.