Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2011 / 5

An isolated singularity at $a$ means that the function is holomorphic on some punctured disc $0<|z-a|<r$, while it is undefined or fails to be holomorphic at $a$. The punctured neighbourhood contains no other singularities; this isolation is needed before applying the usual removable/pole/essential classification.

A removable singularity at $a$ is an isolated singularity for which one can assign a value at $a$ to obtain a holomorphic function on the whole disc. Equivalently its Laurent series has no negative powers. A finite limit $L$ supplies the extension value $f(a)=L$.

A pole of order $m\ge 1$ at $a$ meansEquivalently the lowest nonzero power in its Laurent series is $(z-a{)}^{-m}$. In particular, $(z-a{)}^{m}f(z)$ tends to a finite nonzero limit.

The possible singular points are $i,-i,1$. Near $i$, put $h(z)=1/((z+i)(z-1{)}^2)$; thenThe numerator vanishes at $i$, so the apparent pole cancels and $i$ is removable. Its extension value is $h^{\prime }(i)=1/4+3i/8$, obtained from $h^{\prime }/h=-1/(z+i)-2/(z-1)$.

At $-i$, the first term has a simple pole with nonzero residueThe subtracted term is holomorphic there, so $-i$ is a simple pole. At $1$,so $1$ is a double pole. There are no other finite singularities.

Away from zero, singularities occur exactly when $e^{1/z}=-1$, namelyFor $D(z)=z(e^{1/z}+1)$, differentiation at $z_k$ gives $D^{\prime }(z_k)=1/z_k\ne 0$. Thus every $z_k$ is a simple pole, with residue $z_k$.

As $|k|\to \infty$, these poles accumulate at zero. Therefore zero is a non-isolated singularity. Every punctured disc around it contains poles, so the function is not holomorphic on any punctured disc there. In particular zero cannot be classified as an isolated essential singularity; the isolated-singularity classification does not apply. The displayed poles and their accumulation point zero exhaust the finite singularities.