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

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.