Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 5

The Laurent theorem states that a holomorphic function on the annulus $R<|z-a|<S$ has a unique Laurent seriesThe integration circle is counterclockwise. The series converges absolutely at every point of the annulus and uniformly on every compact subannulus. The coefficient integrals are independent of the choice of $\rho$ by the Cauchy integral theorem applied between two such circles. These statements include $R=0$ or $S=\infty$ when the indicated annulus has those endpoints.

An isolated singularity at $z_0$ means that some radius $r>0$ has $f$ holomorphic on the punctured disk $0<|z-z_0|<r$, while $f$ is not initially defined as a holomorphic function at $z_0$. The adjective isolated refers to the absence of other singularities in this punctured neighbourhood; it permits a removable singularity.

The point is a removable singularity if assigning a value to $f(z_0)$ makes $f$ holomorphic on a full neighbourhood of $z_0$. Equivalently, its Laurent series has no nonzero coefficients of negative powers. The assigned value is its constant coefficient, or equivalently the finite limit ${\lim }_{z\to z_0}f(z)$.

A pole of order $n\ge 1$ meansEquivalently, the Laurent series has a finite nonzero principal part, with lowest power exactly $(z-z_0{)}^{-n}$. Thus $(z-z_0{)}^{n}f(z)$ extends holomorphically with nonzero value at $z_0$; using a smaller positive power would not remove the singularity.

An essential singularity is an isolated singularity that is neither a removable singularity nor a pole. In its Laurent series, infinitely many coefficients of negative powers are nonzero. The alternatives follow because no negative terms give a removable singularity, while finitely many with a lowest nonzero power give a pole.

Inside the unit circle, use the geometric series for $1/(1-z)$:Outside the unit circle, expand in powers of $1/z$ instead:The different Laurent series reflect different annuli separated by the pole at one. Both follow from absolutely convergent geometric series in their stated regions.

Take the Laurent series of $f$ on $0<|z|<r$. For every integer $k\ge 1$, its coefficient of $z^{-k}$ satisfies, on a circle of radius $0<\rho <r$,The coefficient is independent of $\rho$. Letting $\rho \downarrow 0$ therefore gives $c_{-k}=0$ for every $k\ge 1$. The Laurent series is a power series with no negative powers and supplies a holomorphic extension at zero. Thus zero is a removable singularity. This is the Riemann removable singularity theorem, with boundedness used to eliminate the entire principal part.

Suppose, towards a contradiction, that for some $r,w,\epsilon$ the punctured-disk image misses the open disk of radius $\epsilon$ about $w$. Thenis holomorphic and bounded by $1/\epsilon$ on $0<|z|<r$. Part (i) extends it to a holomorphic function $H$ on the full disk. If $H(0)\ne 0$, then $w+1/H$ extends $f$ holomorphically at zero, making zero a removable singularity.

If $H(0)=0$, the function $H$ is not identically zero, since it is nonzero on the punctured disk. The Taylor series and identity theorem give a finite zero order $k\ge 1$, so $H(z)=z^{k}q(z)$ with $q$ holomorphic and $q(0)\ne 0$. Hencehas a pole of order $k$. Both alternatives contradict an essential singularity. ConsequentlyThis is the Casorati-Weierstrass theorem: every complex value can be approximated arbitrarily closely in every punctured neighbourhood of an essential singularity.