Past exam of the short options paper of the physics Prelims of the University of Oxford / 2021 / S01 / 1

The apparent singularity at zero cancels: the Taylor series of the sine givesThus zero is a removable singularity, with extension $f(0)=1/2$. At $z=-2$, the other terms are holomorphic, whereas $(z+2{)}^{-1}$ has a simple pole. ThereforeThere are no other finite singularities. If “all” includes the point at infinity on the Riemann sphere, it is an essential singularity: $f(1/w)=w/(1+2w)+w^2\sin (1/w)-w$ has infinitely many negative powers in its Laurent series at $w=0$.

The geometric series for $(2+z{)}^{-1}$ converges for $|z|<2$, while the cancelled sine terms have an everywhere convergent Taylor series. Hence the requested Laurent series has no negative powers:For example,The extension through zero also makes this a Taylor series. The pole at $-2$ fixes its radius of convergence at two.

Assume positive, counterclockwise orientation of a contour. Factor the polynomial denominator asOnly $z=\pm (\sqrt{3}-\sqrt{2})$ lie inside the unit circle. These are simple poles. Writing the denominator as $P(z)$, their residues areThe residue theorem therefore givesReversing orientation of a contour changes the sign.

Parametrize the positively oriented unit circle by $z=e^{i\theta }$. Then $dz=izd\theta$ andConsequently the contour integral already evaluated satisfies