# 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 $$r < s < t < R$$.

## Coefficients

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

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

## Singularities

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 $$k<0$$

### Removable

An isolated singularity $$z_0$$ of $$f$$ is removable if and only if $$f$$ is bounded near $$z_0$$.

### Pole

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

### Essential

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 $$z_0$$,

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

## References

[1]
Petra Bonfert-Taylor. Introduction to complex analysis. Course on Coursera.