Complex Analysis/Residue Theorem

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

Let ${\displaystyle f}$ be a holomorphic function in a region ${\displaystyle G}$ except for a discrete set of isolated singularities ${\displaystyle S\subseteq G}$, and let ${\displaystyle \Gamma }$ be a null-homologous Chain in ${\displaystyle G}$ that does not intersect any point of ${\displaystyle S}$. Then, the following holds:

${\displaystyle \int _{\Gamma }f(z)\,dz=2\pi i\cdot \sum _{z\in S}n(\Gamma ,z)\cdot \mathrm {res} _{z}(f).}$

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

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

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

For the singularities ${\displaystyle z_{i}\in S}$ with ${\displaystyle \mu (\Gamma ,z_{i})\not =0}$ and ${\displaystyle 1\leq i\leq k}$, let

${\displaystyle h_{i}(z)=\sum _{j=-1}^{-\infty }a_{ij}(z-z_{i})^{j}}$

be the main part of the Laurent Expansion of ${\displaystyle f}$ around ${\displaystyle z_{i}}$. The function ${\displaystyle h_{i}}$ is holomorphic on ${\displaystyle \mathbb {C} \setminus \{z_{i}\}}$.

Subtracting all the principal parts ${\displaystyle h_{i}}$ corresponding to ${\displaystyle z_{i}}$ from the given function ${\displaystyle f}$, we obtain

${\displaystyle g:=f-\sum _{i=1}^{k}h_{i}}$

a function on ${\displaystyle G\setminus T}$ that now has only removable singularities.

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

By the Cauchy Integral Theorem for ${\displaystyle \Gamma }$, we have

${\displaystyle \int _{\Gamma }g(z)\,dz=0}$

so, by the definition of ${\displaystyle g}$,

${\displaystyle {\begin{array}{rcl}\displaystyle \int _{\Gamma }f(z)\,dz&=&\displaystyle \int _{\Gamma }g(z)\,dz+\sum _{i=1}^{k}\int _{\Gamma }h_{i}(z)\,dz\\&=&\displaystyle \sum _{i=1}^{k}\int _{\Gamma }h_{i}(z)\,dz.\end{array}}}$

The computation of the integral over ${\displaystyle f}$ reduces to computing the integrals of the principal parts ${\displaystyle h_{i}}$ for ${\displaystyle 1\leq i\leq k}$. Using the linearity of the integral, we have:

${\displaystyle {\begin{array}{rcl}\displaystyle \int _{\Gamma }h_{i}(z)\,dz&=&\displaystyle \sum _{j=-\infty }^{-i}a_{ij}\int _{\Gamma }(z-z_{i})^{j}\end{array}}}$

the terms ${\displaystyle \displaystyle \int _{\Gamma }(z-z_{i})^{j}}$ For ${\displaystyle j\leq -2}$, have antiderivatives, so ${\displaystyle \displaystyle \int _{\Gamma }(z-z_{i})^{j}=0}$.

Finally, the computation of the integrals of the principal parts yields, using the definition of the Winding number:

${\displaystyle {\begin{array}{rcl}\displaystyle \int _{\Gamma }h_{i}(z)\,dz&=&\displaystyle \sum _{j=-\infty }^{-i}a_{ij}\int _{\Gamma }(z-z_{i})^{j}\\&=&\displaystyle a_{i,-1}\int _{\Gamma }(z-z_{i})^{-1}\,dz\\&=&\displaystyle a_{i,-1}\cdot 2\pi \cdot i\cdot n(\Gamma ,z_{i})\\&=&\displaystyle 2\pi \cdot i\cdot n(\Gamma ,z_{i})\cdot \mathrm {res} _{z_{i}}f\end{array}}}$

after.

Thus, the statement follows as:

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

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

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

• null-homologous
• Isolated singularity

You can display this page as Wiki2Reveal slides

The Wiki2Reveal slides were created for the Complex Analysis' and the Link for the Wiki2Reveal Slides was created with the link generator.

• This page is designed as a PanDocElectron-SLIDE document type.
• Source: Wikiversity https://en.wikiversity.org/wiki/Complex%20Analysis/Residue%20Theorem
• see Wiki2Reveal for the functionality of Wiki2Reveal.

This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:

• Source: Kurs:Funktionentheorie/Residuensatz - URL:

https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Residuensatz

• Date: 01/05/2024