Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 16B (Complex Methods)

The residue is the coefficient of $(z-a{)}^{-1}$ in the Laurent series:Equivalently it is $\frac{1}{2\pi i}{\int }_{|z-a|=r}f(z)dz$ for a positively oriented circle small enough to enclose no other singularities.

If the pole has order $k+1$, multiplying the Laurent series by $(z-a{)}^{k+1}$ gives a function $h$ with a holomorphic extension across $a$:Its coefficient of $(z-a{)}^k$ is exactly $c_{-1}$. The Taylor series coefficient formula gives $h^{(k)}(a)=k!c_{-1}$. Continuity of this derivative at $a$ therefore proves

Integrate $F(z)=(1+z^2{)}^{-(k+1)}$ along the interval $[-R,R]$ closed by the positively oriented upper semicircle, with $R>1$. There is just one enclosed pole, at $i$, of order $k+1$. There,Differentiating $k$ times givesBy part (b),On the semicircle $|1+z^2|\ge R^2-1$, so its integral has absolute value at most $\pi R/(R^2-1{)}^{k+1}$, tending to zero. The real integral converges absolutely, so the residue theorem and the limit $R\to \infty$ give