Laurent series

Personal notes on Laurent series in complex analysis.

Notes taken while taking a course [1].

Laurent series

If \(f:U \rightarrow \Complex\) is analytic and \(D = \{r < |z-z_0| < R\} \subset U\), then there is a sequence \(a_k\) where \(k \in \mathbb{Z}\) such that

\[ f(z) = \sum_{k=-\infty}^\infty a_k (z-z_0)^k \]

This is a Laurent series expansion for \(f\). It converges in \(D\), and it converges absolutely and uniformly for any \(\{s < |z-z_0| < t\}\) where \(r < s < t < R\).


If \(f\) is analytic in \(\{ r < | z - z_0 | < R \}\), and \(a_k\) is such that

\[ a_k = \frac{1}{2 \pi i} \int_{|z-z_0|=s} \frac{f(z)}{(z-z_0)^{k+1}} \, dz \]

for any \(s\) where \(r < s < R\), then

\[ f(z) = \sum_{k=-\infty}^\infty a_k (z-z_0)^k \]


A singularity \(z_0\) is an isolated singularity of \(f\) if \(f\) is analytic in \(\{ 0 < | z - z_0 | < r \}\) for some \(r>0\).

If \(z_0\) is an isolated singularity for \(f(z) = \sum_{k=-\infty}^\infty a_k (z-z_0)^k\), then the singularity \(z_0\) is

  • removable if \(a_k=0\) for all \(k<0\)
  • a pole if there is some \(N\) such that \(a_N \neq 0\) and \(a_k = 0\) for all \(k < -N\); \(N\) is the order of the pole
  • essential if \(a_k \neq 0\) for infinitely many \(k<0\)


An isolated singularity \(z_0\) of \(f\) is removable if and only if \(f\) is bounded near \(z_0\).


An isolated singularity \(z_0\) of \(f\) is a pole if and only if \(\lim_{z \rightarrow z_0} |f(z)| = \infty\).


If \(z_0\) is an essential singularity of \(f\), then for every \(w \in \Complex\) there exists a sequence \(\{z_n\}\) such that

\[ \lim_{n \rightarrow \infty} z_n = z_0 \]

\[ \lim_{n \rightarrow \infty} f(z_n) = w \]

In addition, for every \(w \in \Complex\) with at most one exception, there exists a sequence \(\{z_n\}\) such that

\[ \lim_{n \rightarrow \infty} z_n = z_0 \]

\[ f(z_n) = w \]

Residue theorem

If these are the case:

  • \(D\) is a simply connected domain
  • \(f\) is analytic in \(D\) except for isolated singularities
  • \(C\) is a simple closed curve oriented counterclockwise with no singularities of \(f\)
  • \(z_1\), …, \(z_n\) are the isolated singularities of \(f\) which lie inside \(C\)

then the following holds:

\[ \int_C f(z) \, dz = 2 \pi i \sum_{k=1}^n \mathrm{Res}(f, z_k) \]

If \(z_0\) is an isolated singularity of \(f\), the residue of \(f\) at \(z_0\) is \(a_{-1}\), where \(f(z) = \sum_{k=-\infty}^\infty a_k (z - z_0)^k\).

\[ \mathrm{Res}(f, z_0) = a_{-1} \]

If an isolated singularity \(z_0\) of \(f\) is removable,

\[ \mathrm{Res}(f, z_0) = 0 \]

If \(z_0\) is a simple pole,

\[ \mathrm{Res}(f, z_0) = \lim_{z \rightarrow z_0} (z - z_0) f(z) \]

More generally, if \(z_0\) is a pole of order \(n\),

\[ \mathrm{Res}(f, z_0) = \frac{1}{(n-1)!} \lim_{z \rightarrow z_0} \frac{d^{n-1}}{dz^{n-1}} \left( (z-z_0)^n f(z) \right) \]

If \(f(z) = \frac{g(z)}{h(z)}\), where \(g\) and \(h\) are analytic near \(z_0\) and \(h\) has a simple zero at \(z_0\),

\[ \mathrm{Res}(f, z_0) = \frac{g(z_0)}{\frac{dh}{dz}(z_0)} \]

See also


Petra Bonfert-Taylor. Introduction to complex analysis. Course on Coursera.