Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 3 / 17E (Further Analysis)

Let $K\subset \mathbb{C}\setminus \mathbb{Z}$ be compact and contained in $|z|\le R$. For $|n|>2R$, $|z-n|\ge |n|/2$, hence $|(z-n{)}^{-2}|\le 4/n^2$ uniformly on $K$. The tail is therefore absolutely and uniformly convergent by the Weierstrass M-test. Its finitely many remaining summands are holomorphic on a neighborhood of $K$.

The theorem that a locally uniform limit of holomorphic functions is holomorphic now proves that the sum is analytic on $\mathbb{C}\setminus \mathbb{Z}$. Absolute convergence also permits a shift of the integer index:Thus the function has period one.

Fix $z_0\ne 0,1$ and take a small disk about it that avoids zero and one. There is a holomorphic logarithm $L$ on this disk. The function $L(z)/(2\pi i)$ cannot be an integer there: if it were an integer, exponentiation would give $z=1$. Thus $f(L(z)/(2\pi i))$ is holomorphic on the disk.

Where the prescribed principal logarithm and $L$ are both evaluated, they differ by $2\pi ik$ for an integer $k$. By the period-one identity, their compositions with $f$ coincide. This remains true on either side of the principal branch cut, and at a point on that cut one may instead choose the logarithm with argument between zero and $2\pi$. The local holomorphic expressions therefore agree with the stated pointwise definition and with each other. This is periodic holomorphic descent through the exponential map, provingThe principal logarithm itself is not analytic across its cut; periodicity of the outer function removes that obstruction.

Near zero the $n=0$ summand supplies the pole and the remaining series is holomorphic, so $f(w)=w^{-2}+F(w)$ with $F$ holomorphic near zero. Choose the local logarithm with $L(1)=0$. Writing $h=z-1$ gives $L(z)=h-h^2/2+O(h^3)$ and henceTherefore $z=1$ is a double pole, with leading coefficient ${\lim }_{z\to 1}(z-1{)}^{2}g(z)=-4{\pi }^2$. Its residue, if desired, is $-4{\pi }^2$.