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==Definition==
Let <math>G \subseteq \mathbb{C}</math> be a domain, <math>z_0 \in G</math>, and let <math>f</math> be holomorphic except for isolated singularities <math>S \subset G</math>, i.e., <math>f\colon G \setminus S \to \mathbb{C}</math> is holomorphic. If <math>z_0 \in S \subset G</math> is an [[Complex Analysis/Isolated singularity|Isolated singularity]] of <math>f</math> with <math>D_r(z_0) \cap S = {z_0}</math>, the residue is defined as:

:<math>\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</math>.

== Relation Between Residue and Laurent Series ==
If <math>f</math> is represented around an isolated singularity <math>z_0 \in S \subset G</math> as a Laurent series, the residue can be computed as follows.
With <math>f(z) = \sum_{n=-\infty}^\infty a_n (z-z_0)^n</math> as the [[Laurent Series|Laurent Series]] expansion of <math>f</math> around <math>z_0</math>, it holds that:
:<math>\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}</math>.
It must be taken into account that the closed disk <math>\overline{D_r(z_0)}</math> contains only the singularity <math>z_0 \in S </math>, i.e<math>\overline{D_r(z_0)} \cap s = \{z_0\}</math>.

Thus, one can read off the 'residue' <math>\mathrm{res}_{z_0}(f) = a_{-1}</math> from the Laurent expansion of  around  at the  <math>f</math> um <math>z_0</math> an -1-ten coefficient of .
==Considerations==
The closed disk <math>\overline{D_r(z_0)}</math> must contain only the singularity <math>z_0 \in S</math>, meaning<math>\overline{D_r(z_0)} \cap S = {z_0}</math>.
Thus, the residue <math>\mathrm{res}{z_0}(f) = a{-1}</math> can be read off directly as the coefficient of <math>(z-z_0)^{-1}</math> in the Laurent series expansion of <math>f</math> around <math>z_0</math>.
==Etymology==
The term "residue" (from Latin residuere – to remain) is used because in integration along the path <math>\gamma(t) := z_0 + r\cdot e^{it}</math> with <math>t \in [0, 2\pi]</math> around the circle centered at <math>z_0</math>, the following holds:

<center><math>
\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}
</math></center>  
Thus, the residue is what remains after integrating.  
==Computation for Poles==
If <math>z_0 \in U</math> is a pole of order <math>m</math> of <math>f</math>, the [[Laurent Series|Laurent Series]] expansion of <math>f</math> around <math>z_0</math> has the form:
:<math>f(z) = \sum_{k=-m}^\infty a_k (z-z_0)^k</math>
with <math>a_{-m} \neq 0</math>.

===Proof 1: Removing Principal Part by Multiplication===
By multiplying with <math>(z-z_0)^m</math>, we get:
:<math>g_m(z) := (z-z_0)^m \cdot f(z) = \sum_{k=0}^\infty a_{k-m} (z-z_0)^k.</math>
The residue <math>a_{-1}</math> is then the coefficient of <math>(z-z_0)^{m-1}</math> in the power series of <math>g_m(z)</math>.

===Proof 2: Using (m-1)-fold Differentiation===
By differentiating <math>m-1</math> times, the first <math>m-1</math> terms from <math>n=0</math> to <math>n=m-2</math> vanish. The residue is then found as the coefficient of <math>(z-z_0)^0</math> in:
:<math>g_m^{(m-1)}(z) = \sum_{k=m-1}^\infty \frac{k!}{(k-m+1)!} a_{k-m}(z-z_0)^{k-m+1}.</math>

===Proof 3: Limit Process to Compute Coefficient of <math>(z-z_0)^0</math>===
By shifting the index:
:<math>g_m^{(m-1)}(z) = \sum_{k=-1}^\infty \frac{(m+k)!}{(k+1)!} a_k (z-z_0)^{k+1}.</math>
Taking the limit <math>z \to z_0</math>, all terms with <math>k \geq 0</math> vanish, leaving:
:<math>\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}.</math>
Thus, the residue can be computed <math>z \to z_0</math> using:
:<math>\mathrm{res}{z_0}(f) = a{-1} = \frac{1}{(m-1)!} \cdot \lim_{z \to z_0} g_m^{(m-1)}(z).</math>

==Exercises for Students==

*Explain why, in the Laurent series expansion, all terms from the principal and outer parts, i.e., <math>n \in \mathbb{Z}</math> with <math>n \neq -1</math>, yield integrals that evaluate to zero:
:<math>\int_{\partial D_r(z_0)} a_n \cdot (\xi - z_0)^n, d\xi = 0.</math>

*Why can the order of integration and series expansion be interchanged?
::<math>\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).</math>

*Given the function <math>f:\mathbb{C}\setminus {i} \to \mathbb{C}</math> with <math>z \mapsto f(z) = \frac{e^{z-i}}{(z-i)^5}</math>, calculate the residue <math>\mathrm{res}_{z_0}(f)</math> at <math>z_0 := i</math>.

==See Also==

*[[Residue|Residue]]
*[[Complex_analysis/Examples_of_Laurent_series|Examples of Laurent Series]]
== Page Information ==
=== Translation and Version Control ===

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* Source: [[v:de:Kurs:Funktionentheorie/Residuum|Kurs:Funktionentheorie/Residuum ]] - URL:
https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Residuum 
* Date: 12/26/2024<span type="translate" src="Kurs:Funktionentheorie/Residuum " srclang="de" date="12/26/2024" time="11:28" status="inprogress"></span>

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