Home | Assessment | Notes | Worksheets | Blackboard
Taylor's theorem gives us a clear picture of the local structure of a holomorphic function: if $f$ is holomorphic on $\ball(b,R)$ then $f$ can be represented by a power series centered at $b$ with radius of convergence at least $R$. Laurent's theorem augments this with a clear picture of structure of a holomorphic function on an annulus: if $f$ is holomorphic on $\Ann(b,r,R)$ then $f$ can be represented by a Laurent series on $\Ann(b,r,R)$. We will now concentrate on the case $r = 0$. That is, we will say a bit more about the structure of a holomorphic function on a punctured ball $\Ann(b,0,R)$.
An isolated singularity of a function $f$ is a point $b \in \C$ such that $f$ is holomorphic on $\Ann(b,0,R)$ for some $R > 0$.
If $b$ is an isolated singularity of $f$ then $f$ is holomorphic on $\Ann(b,0,R)$ so can be represented by a Laurent series \[ f(z) = \sum_{n=1}^\infty a_n (z-b)^{-n} + a_0 + \sum_{n=1}^\infty a_n (z-b)^n \] on $\Ann(b,0,R)$. We classify the nature of the singularity $b$ according to the principal term. Each of the following are possible.
The point $b$ is then described respectively as follows.
If $b$ is an isolated singularity that is removable then $f$ is equal to the power series \[ g(z) = a_0 + \sum_{n=1}^\infty a_n (z-b)^n \] on $\Ann(b,0,R)$. But, as a power series on $\ball(b,R)$ the function $g$ is continuous and differentiable at $b$. This is the sense in which the singularity is removable: if we extend the definition of $f$ to $\ball(b,R)$ by declaring that $f(b) = a_0$ we get a function that is holomorphic on $\ball(b,R)$.
Define $f : \C \setminus \{0\} \to \C$ by \[ f(z) = \dfrac{\sin(z)}{z} \] for all $z \ne 0$. It is not defiend at $0$ and therefore certainly not holomorphic at $0$, but is differentiable at every non-zero complex number, so $0$ is an isolated singularity of $f$. If $z \ne 0$ then \[ \begin{align*} \dfrac{\sin(z)}{z} & = \dfrac{1}{z} \sum_{k=0}^\infty \dfrac{(-1)^k}{(2k+1)!} z^{2k+1} \\ & = \sum_{k=0}^\infty \dfrac{(-1)^k}{(2k+1)!} z^{2k} \\ & = 1 - \dfrac{z^2}{3!} + \dfrac{z^4}{5!} - \dfrac{z^6}{7!} + \cdots \end{align*} \] which is a power series with infinite radius of convergence. Thus $0$ is a removable singularity of $f$. If we declare $f(0) = 1$ then $f$ is defined on all of $\C$ and is holomorphic there.
When $b$ is an isolated singularity that is a pole we represent $f$ as \[ f(z) = \dfrac{a_{-m}}{(z-b)^m} + \cdots + \dfrac{a_{-1}}{(z-b)} + a_0 + \sum_{n=1}^\infty a_n (z-b)^n \] on $\Ann(b,0,R)$. We say that $f$ has a pole of order $m$ at $b$ in this case. A pole of order 1 is called a simple pole.
Define $f : \C \setminus \{0\} \to \C$ by \[ f(z) = \dfrac{\sin(z)}{z^4} \] for all $z \ne 0$. If $z \ne 0$ we have \[ \begin{align*} f(z) & = \frac{1}{z^4} \left( z - \dfrac{z^3}{3!} + \dfrac{z^5}{5!} - \dfrac{z^7}{7!} + \cdots \right) \\ & = \dfrac{1}{z^3} - \dfrac{1}{3!} \dfrac{1}{z} + \dfrac{z}{5!} - \dfrac{z^3}{7!} + \cdots \end{align*} \] and read off that $f$ has a pole of order $3$ at $0$.
Often we will be interested in functions $f : D \to \C$ that are holomorphic on $D \setminus \{b_1,\dots,b_t \}$ and have poles at each of the singularities $b_1,\dots,b_t$.
By a meromorphic function on a domain $D$ we will mean a function $f : D \setminus \{ b_1,\dots,b_t \}$ with poles at each of the points $b_1,\dots,b_t$.
When $b$ is an isolated singularity that is an essential singularity we represent $f$ as \[ f(z) = \sum_{n=1}^\infty a_n (z-b)^{-n} + a_0 + \sum_{n=1}^\infty a_n (z-b)^n \] on $\Ann(b,0,R)$ with infinitely many of the coefficients $a_{-n}$ non-zero.
The functions \[ \exp(1/z^2) = 1 + \dfrac{1}{z^2} + \dfrac{1}{2!} \dfrac{1}{z^4} + \dfrac{1}{3!} \dfrac{1}{z^6} + \cdots \] and \[ \sin(1/z) = \dfrac{1}{z} - \dfrac{1}{3!} \dfrac{1}{z^3} + \dfrac{1}{5!} \dfrac{1}{z^5} - \dfrac{1}{7!} \dfrac{1}{z^7} + \cdots \] both have essential singularities at $0$.
One remarkable result about essential singularities is that the image of $\Ann(b,0,R)$ under $f$ is the punctured plane for each $R > 0$ whenever $f$ has an essential singularity at $b$. (This is called the great Picard theorem.) Essential singularities are difficult to handle and we will say no more about them in this course.