# Complex Analysis/Residuals
## Definition

Let $G\subseteq \mathbb {C}$ be a domain, $z_{0}\in G$, and $f$ a function that is holomorphic except for isolated singularities $S\subset G$, i.e., $f:G\setminus S\to \mathbb {C}$ is holomorphic. If $z_{0}\in S\subset G$ is an Isolated singularity of $f$ with $D_{r}(z_{0})\cap S={z_{0}}$, the residue is defined as:
    $\mathrm {res} {z_{0}}(f):={\frac {1}{2\pi i}}\int {\partial D_{r}(z_{0})}f(\xi ),d\xi ={\frac {1}{2\pi i}}\int _{|\xi -z_{0}|=r}f(\xi ),d\xi$.

## Relation between Residue and Laurent Series

If $f$ is expressed as a Laurent series around an isolated singularity $z_{0}\in S\subset G$, the residue can be computed as follows: With $f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n}$ as the Laurent Expansion of $f$ around $z_{0}$, it holds:
    $\mathrm {res} {z_{0}}(f)={\frac {1}{2\pi i}}\int {\partial D_{r}(z_{0})}f(\xi ),d\xi ={\frac {1}{2\pi i}}\int _{\partial D_{r}(z_{0})}a_{-1}\cdot (\xi -z_{0})^{-1},d\xi ={\frac {1}{2\pi i}}a_{-1}\cdot \underbrace {\int _{\partial D_{r}(z_{0})}(\xi -z_{0})^{-1},d\xi } {=2\pi i}=a{-1}$.
Here, it is assumed that the closed disk ${\overline {D_{r}(z_{0})}}$ contains only the singularity $z_{0}\in S$, i.e., ${\overline {D_{r}(z_{0})}}\cap S={z_{0}}$.
Thus, the residue $\mathrm {res} {z_{0}}(f)=a{-1}$ can be identified as the coefficient of $(z-z_{0})^{-1}$ in the Laurent series of $f$ around $z_{0}$.

## Terminology

The residue (from Latin residuere - to remain) is named so because, during integration along the path $\gamma (t):=z_{0}+r\cdot e^{it}$ with $t\in [0,2\pi ]$ around $z_{0}$, the following holds:
${\begin{array}{rl}\displaystyle \int _{|w-z_{0}|=r}f(w)\,dw&=\displaystyle \sum _{n=-\infty }^{+\infty }a_{n}\int _{|w-z_{0}|=r}(w-z_{0})^{n}\,dw\\&=2\pi i\cdot a_{-1}\end{array}}$
The residue is, therefore, what "remains" after integration.

## Calculation for Poles

If $z_{0}\in U$ is a pole of order $m$ of $f$, the Laurent Expansion of $f$ around $z_{0}$ has the form:
    $f(z)=\sum _{k=-m}^{\infty }a_{k}(z-z_{0})^{k}$
with $a_{-m}\neq 0$.

### Proof 1: Removing the principal part by multiplication

By multiplying with $(z-z_{0})^{m}$, we obtain:
    $g_{m}(z):=(z-z_{0})^{m}\cdot f(z)=\sum _{k=0}^{\infty }a_{k-m}(z-z_{0})^{k}$
The residue $a_{-1}$ is now the coefficient of $(z-z_{0})^{m-1}$ in the power series of $g_{m}(z)$.

### Proof 2: Using (m-1)-fold differentiation

Through $m-1$-fold differentiation, the first $m-1$ terms in the series, from $n=0$ to $m-2$, vanish. The residue is then the coefficient of $(z-z_{0})^{0}$, yielding:
    $g_{m}^{(m-1)}(z)=\sum _{k=m-1}^{\infty }{\frac {k!}{(k-m+1)!}}a_{k-m}(z-z_{0})^{k-m+1}$.

### Proof 3: Limit process to find the coefficient of $(z-z_{0})^{0}$

By shifting the index, we obtain:
    $g_{m}^{(m-1)}(z)=\sum _{k=-1}^{\infty }{\frac {m+k!}{(k+1)!}}a_{k}(z-z_{0})^{k+1}$
Taking the limit $z\to z_{0}$, all terms with $k\geq 0$ vanish, yielding:
    $\lim _{z\to z_{0}}g_{m}^{(m-1)}(z)={\frac {(m-1)!}{0!}}\cdot a_{-1}\cdot (z-z_{0})^{0}=(m-1)!\cdot a_{-1}$.
Thus, the residue can be computed using the limit $z\to z_{0}$:
    $\mathrm {res} {z_{0}}(f)=a{-1}={\frac {1}{(m-1)!}}\cdot \lim _{z\to z_{0}}g_{m}^{(m-1)}(z)$.

## Tasks for Students

* Explain why, during integration of the Laurent series, all terms from the regular part and all terms with index $n\in \mathbb {Z}$ with $n\neq -1$ contribute
    $\int _{\partial D_{r}(z_{0})}a_{n}\cdot (\xi -z_{0})^{n},d\xi =0$.
* Why is it allowed to interchange the processes of integration and series expansion?
    $\sum _{n=-\infty }^{+\infty }\int _{\partial D_{r}(z_{0})}a_{n}(\xi -z_{0})^{n}d\xi =\int _{\partial D_{r}(z_{0})}\sum _{n=-\infty }^{+\infty }a_{n}(\xi -z_{0})^{n}d\xi ={\frac {1}{2\pi i}}\int _{\partial D_{r}(z_{0})}f(\xi ),d\xi =\mathrm {res} _{z_{0}}(f)$
* Given the function $f:\mathbb {C} \setminus {i}\to \mathbb {C}$ with $z\mapsto f(z)={\frac {e^{z-i}}{(z-i)^{5}}}$, compute the residue $\mathrm {res} _{z_{0}}(f)$ with $z_{0}:=i$!.

## See Also

* Residuals
* Laurent series examples

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