Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 4 / a

A function has an isolated singularity at $a$ if, for some $r>0$, it is defined and holomorphic on the punctured disc $0<|z-a|<r$, while a holomorphic extension through $a$ has not been assumed. In particular, its domain omits $a$, or its assigned value there fails to give such an extension. The key local condition isThe convention includes removable singularities, even when a suitable extension would eliminate the missing point.

An isolated singularity at $a$ is a removable singularity when there is a holomorphic function $F$ on a full disc $B(a,r)$ agreeing with $f$ away from $a$. Equivalently,The Riemann removable singularity theorem also characterizes this by boundedness of $f$ on a sufficiently small punctured disc. A preassigned incorrect value at $a$ is replaced by $L$.

A holomorphic function has a zero of order $n$ at $a$ when it is holomorphic on a neighbourhood of $a$ and factors asEquivalently, $f(a)=f^{\prime }(a)=\cdots =f^{(n-1)}(a)=0$ and $f^{(n)}(a)\ne 0$. The order of a zero of a holomorphic function is the degree of the first nonzero term of its Taylor series.

A function has a pole of order $n$ at $a$ when, on a punctured neighbourhood,Equivalently, $(z-a{)}^{n}f(z)$ has a removable singularity whose extension is nonzero at $a$. After shrinking the neighbourhood so $h\ne 0$, its reciprocal $1/f$ extends with a zero of order $n$. In particular, $|f(z)|\to \infty$ at a pole.