Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 / Analysis / 5

For any $0<r<1$, a Laurent series converges uniformly on the circle $|z|=r$, so termwise contour integration is valid. Multiplication by $z^{-k-1}$ and integration gives, for every integer $k$,Indeed, the integral of $z^{j-k-1}$ is $2\pi i$ for $j=k$ and zero otherwise. These coefficients depend only on $f$, so any two Laurent expansions must agree coefficient by coefficient. This proves uniqueness, including all terms of the principal part.

A removable singularity means that $f$ has a holomorphic extension to the full disk, assigning a finite value at zero. Equivalently its Laurent series has no negative-power terms. One direction follows by the Taylor expansion of the extension; the other by the convergent nonnegative-power series. Thus $\boxed{a_k=0\text{ for all }k<0}$ is the coefficient criterion.

A pole of order $n$ means $f(z)=z^{-n}g(z)$ with $g$ holomorphic near zero and $g(0)\ne 0$. Equivalently, the principal part is finite and starts with $\boxed{a_{-n}{z}^{-n},a_{-n}\ne 0}$, while $a_k=0$ for $k<-n$. In particular $n$ is the exact order, not merely an upper bound.

An essential singularity is an isolated singularity that is neither removable nor a pole. In terms of the Laurent series this means $\boxed{\text{infinitely many negative-power coefficients are nonzero}}$: no finite multiplication by a power of $z$ removes the singularity.

For $|z|<1$, differentiate the geometric series to obtain $(1-t{)}^{-2}={\sum }_{k\ge 0}(k+1)t^k$. Since $(z-i{)}^2=-(1-z/i{)}^2$,The convergence region is precisely $0<|z|<1$, the specified punctured disk; the other pole is on its boundary at $i$.

If $f=z^{-n}g$ with $g(0)\ne 0$, continuity gives a small disk on which $|g(0)|/2\le |g(z)|\le 3|g(0)|/2$. Multiplying by $|z{|}^{-n}$ gives the two required bounds.

Conversely, set $g(z)=z^{n}f(z)$ on the punctured disk. The upper bound makes $g$ bounded, so the allowed removable-singularity theorem gives a holomorphic extension. The lower bound gives $|g(0)|\ge C_1>0$ by continuity. Therefore $f=z^{-n}g$ has a pole of exact order $n$. The lower bound is essential: the upper bound alone only proves order at most $n$.