Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 1 / 12G (Complex Analysis or Complex Methods)

The Laurent theorem says that if $f$ is analytic on an annulusthen it has a unique Laurent seriesconverging locally uniformly on that annulus, wherefor any positively oriented circle $C$ in the annulus around $a$.

An isolated singularity at $a$ is a point at which $f$ is not analytic although it is analytic on some punctured neighbourhood. It is removable when every $c_n$ with $n<0$ vanishes; it is a pole of order $m$ when $c_{-m}\ne 0$ and $c_n=0$ for $n<-m$; and it is essential when infinitely many negative-index coefficients are nonzero.

For $0<|z|<1$,For $|z|>1$,The coefficients are unique after the annulus is fixed; these expansions differ because they represent the function on different annuli. At zero the first expansion has principal part $-z^{-1}$, so zero is a simple pole with residue $-1$.

Put $g=1/f$. The hypothesis $|f(z)|\to \infty$ gives $g(z)\to 0$ as $z\to a$, so defining $g(a)=0$ makes $g$ continuous on $U$ and analytic there by the stated assumption. Its zero at $a$ has some finite order $m\ge 1$, and henceThereforehas a pole of order $m$: its Laurent series has $c_{-m}\ne 0$ and $c_n=0$ for $n<-m$.

Now let $f$ be entire and tend to infinity at infinity. The functiontends to infinity as $z\to 0$, so the preceding argument says that $h$ has a pole at zero. If the Taylor series of $f$ is $f(w)={\sum }_{n\ge 0}{a}_n{w}^n$, thenA pole has only finitely many negative powers, so $a_n=0$ for all sufficiently large $n$. Thus $f$ is a polynomial.

Forboth $e^z-1$ and $\log (1+z)$ have a simple zero at zero. Hence $g$ has a simple pole, and its residue is

For $h(z)=\sin z\sin (1/z)$, multiplication of the two Laurent series shows that all powers are even:There are infinitely many negative powers, so zero is an essential singularity. There is no $z^{-1}$ term, and therefore