# Complex Analysis/Residue Theorem

The residue theorem states, how to calculate the integral of a holomorphic function using its Residuals .

## Statement

Let $f$ be a holomorphic function in a region $G$ except for a discrete set of isolated singularities $S\subseteq G$, and let $\Gamma$ be a null-homologous Chain in $G$ that does not intersect any point of $S$. Then, the following holds:
$\int _{\Gamma }f(z)\,dz=2\pi i\cdot \sum _{z\in S}n(\Gamma ,z)\cdot \mathrm {res} _{z}(f).$

## Proof

The sum in the statement of the residue theorem is finite because $\Gamma$ can enclose only finitely many points of the discrete set $S$ of singularities.

### Step 1 - Reduction to a finite number of summands

Let $z_{1},\ldots ,z_{k}$ be the points in $S$ for which $n(\Gamma ,z_{i})\neq 0$. The singularities in $S$ that are not enclosed are denoted by $T:={z\in S:n(\Gamma ,z)=0}$.

### Step 2 - Nullhomologous cycle

$\Gamma$ is assumed to be null-homologous in $G$. By the definition of $T$,is $\Gamma$ also null-homologous in $G\setminus T$.

### Step 3 - Principal parts of the Laurent series

For the singularities $z_{i}\in S$ with $\mu (\Gamma ,z_{i})\not =0$ and $1\leq i\leq k$, let
$h_{i}(z)=\sum _{j=-1}^{-\infty }a_{ij}(z-z_{i})^{j}$
be the main part of the Laurent Expansion of $f$ around $z_{i}$. The function $h_{i}$ is holomorphic on $\mathbb {C} \setminus \{z_{i}\}$.

### Step 4 - Subtraction of principal parts

Subtracting all the principal parts $h_{i}$ corresponding to $z_{i}$ from the given function $f$, we obtain
$g:=f-\sum _{i=1}^{k}h_{i}$
a function on $G\setminus T$ that now has only removable singularities.

### Step 5 - Holomorphic extension to $G$

If the singularities $z_{i}$ are Isolated singularity on $G\setminus T$, $g$ can be extended holomorphically to all $z_{1},\ldots ,z_{k}\in G\setminus T$.

### Step 6 - Application of Cauchy's integral theorem

By the Cauchy Integral Theorem for $\Gamma$, we have
$\int _{\Gamma }g(z)\,dz=0$
so, by the definition of $g$,
${\begin{array}{rcl}\displaystyle \int _{\Gamma }f(z)\,dz&=&\displaystyle \int _{\Gamma }g(z)\,dz+\sum _{i=1}^{k}\int _{\Gamma }h_{i}(z)\,dz\\&=&\displaystyle \sum _{i=1}^{k}\int _{\Gamma }h_{i}(z)\,dz.\end{array}}$

### Step 7 - Calculation of integrals of the principal parts

The computation of the integral over $f$ reduces to computing the integrals of the principal parts $h_{i}$ for $1\leq i\leq k$. Using the linearity of the integral, we have:
${\begin{array}{rcl}\displaystyle \int _{\Gamma }h_{i}(z)\,dz&=&\displaystyle \sum _{j=-\infty }^{-i}a_{ij}\int _{\Gamma }(z-z_{i})^{j}\end{array}}$
the terms $\displaystyle \int _{\Gamma }(z-z_{i})^{j}$ For $j\leq -2$, have antiderivatives, so $\displaystyle \int _{\Gamma }(z-z_{i})^{j}=0$.

### Step 8 - Calculation of integrals of the principal parts

Finally, the computation of the integrals of the principal parts yields, using the definition of the Winding number:
${\begin{array}{rcl}\displaystyle \int _{\Gamma }h_{i}(z)\,dz&=&\displaystyle \sum _{j=-\infty }^{-i}a_{ij}\int _{\Gamma }(z-z_{i})^{j}\\&=&\displaystyle a_{i,-1}\int _{\Gamma }(z-z_{i})^{-1}\,dz\\&=&\displaystyle a_{i,-1}\cdot 2\pi \cdot i\cdot n(\Gamma ,z_{i})\\&=&\displaystyle 2\pi \cdot i\cdot n(\Gamma ,z_{i})\cdot \mathrm {res} _{z_{i}}f\end{array}}$
after.

### Step 9 - Calculation of the integrals of the residues

Thus, the statement follows as:
${\begin{array}{rcl}\displaystyle \int _{\Gamma }f(z)\,dz&=&\displaystyle \sum _{i=1}^{k}\int _{\Gamma }h_{i}(z)\,dz\\&=&\displaystyle 2\pi i\sum _{i=1}^{k}n(\Gamma ,z_{i})\mathrm {res} _{z_{i}}f\end{array}}$

## Questions about the residue theorem

* Let $f:G\rightarrow {\widehat {\mathbb {C} }}$ be a meromorphic function (i.e., holomorphic except for a discrete set of singularities in $G$). Why does the cycle $\Gamma$ enclose only finitely many poles?

## Applications

The Zeros and poles counting integral counts the zeros and poles of a meromorphic function.