Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 4

In the Laurent seriesabout a pole of order $m$, the residue is the coefficientwhere the circle is traversed anticlockwise. Equivalently, the residue is $\frac{1}{(m-1)!}{\frac{d^{m-1}}{d{z}^{m-1}}[(z-z_0{)}^{m}f(z)]|}_{z=z_0}$.

The Taylor series of the entire function $g$ givesThe coefficient of $(z-z_0{)}^{-1}$, and therefore the residue, isIf $g(z_0)=0$, the apparent pole is a removable singularity and the residue is zero.

For the simple-contour form of the residue theorem, let $C$ be a positively oriented, simple, closed, piecewise smooth contour. Assume $f$ is holomorphic on a neighbourhood of $C$ and its interior except at finitely many isolated singularities $z_1,\dots ,z_s$, all strictly inside $C$, with none on $C$. ThenThe more general version weights each residue by the contour's winding number, with the usual requirement that the contour be homologous to zero in the domain of analyticity containing it and the nonsingular region.

For the requested application, assume $f$ is holomorphic on a neighbourhood of the contour and its closed interior, and $z_0$ is strictly inside it. Its Taylor series showsOnly the second term contributes a residue. The residue theorem therefore proves the derivative form of the Cauchy integral formula:

Close the real segment $[-R,R]$ by an anticlockwise semicircle in the upper half-plane. For $F(z)=1/[(z^2+1)(z^2+4)]$, the contour integral over the arc is $O(R^{-3})$: its length is $\pi R$, whereas $|z^2+1|\ge R^2-1$ and $|z^2+4|\ge R^2-4$. The enclosed simple poles are $i$ and $2i$, with residuesLetting $R\to \infty$ in the residue theorem yields

Use $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The unit circle is traversed anticlockwise, and the integral becomesOnly the simple pole $z=i/3$ lies inside the contour. Its residue isHence the residue theorem gives