Editing Residue (complex analysis)

{{Short description|Attribute of a mathematical function}}
{{Complex analysis sidebar}}
In [[mathematics]], more specifically [[complex analysis]], the '''residue''' is a [[complex number]] proportional to the [[Contour integration|contour integral]] of a [[meromorphic function]] along a path enclosing one of its [[mathematical singularity|singularities]]. (More generally, residues can be calculated for any function <math> f\colon \mathbb{C} \setminus \{a_k\}_k \rightarrow \mathbb{C}</math> that is [[Holomorphic function|holomorphic]] except at the discrete points {''a''<sub>''k''</sub>}<sub>''k''</sub>, even if some of them are [[essential singularity|essential singularities]].) Residues can be computed quite easily and, once known, allow the determination of general contour integrals via the [[residue theorem]].

== Definition ==
The residue of a [[meromorphic function]] <math>f</math> at an [[isolated singularity]] <math>a</math>, often denoted <math>\operatorname{Res}(f,a)</math>, <math>\operatorname{Res}_a(f)</math>, <math>\mathop{\operatorname{Res}}_{z=a}f(z)</math> or <math>\mathop{\operatorname{res}}_{z=a}f(z)</math>, is the unique value <math>R</math> such that <math>f(z)- R/(z-a)</math> has an [[Analytic function|analytic]] [[antiderivative (complex analysis)|antiderivative]] in a [[punctured disk]] <math>0<\vert z-a\vert<\delta</math>.  

Alternatively, residues can be calculated by finding [[Laurent series]] expansions, and one can define the residue as the coefficient ''a''<sub>−1</sub> of a Laurent series.  

The concept can be used to provide contour integration values of certain contour integral problems considered in the [[residue theorem]]. According to the [[residue theorem]], for a [[meromorphic function]] <math>f</math>, the residue at point <math>a_k</math> is given as: 

: <math>\operatorname{Res}(f,a_k) = {1 \over 2\pi i} \oint_\gamma f(z)\,dz \, .</math>

where <math>\gamma</math> is a [[Curve orientation|positively oriented]] [[Jordan curve|simple closed curve]] around <math>a_k</math> and not including any other singularities on or inside the curve.

The definition of a residue can be generalized to arbitrary [[Riemann surfaces]]. Suppose <math>\omega</math> is a [[One-form|1-form]] on a Riemann surface. Let <math>\omega</math> be meromorphic at some point <math>x</math>, so that we may write <math>\omega</math> in local coordinates as <math>f(z) \; dz</math>. Then, the residue of <math>\omega</math> at <math>x</math> is defined to be the residue of <math>f(z)</math> at the point corresponding to <math>x</math>.

== Contour integration<!-- Section needs introduction --> ==
{{See also|Contour integration}}

=== Contour integral of a monomial ===
Computing the residue of a [[monomial]]

:<math>\oint_C z^k \, dz</math>

makes most residue computations easy to do. Since path integral computations are [[homotopy]] invariant, we will let <math>C</math> be the circle with radius <math>1</math> going counter clockwise. Then, using the change of coordinates <math>z \to e^{i\theta}</math> we find that

: <math>dz \to d(e^{i\theta}) = ie^{i\theta} \, d\theta</math>

hence our integral now reads as

:<math>
\oint_C z^k dz = \int_0^{2\pi} i e^{i(k+1)\theta} \, d\theta
= \begin{cases}
2\pi i & \text{if } k = -1, \\
0 & \text{otherwise}.
\end{cases}
</math>
Thus, the residue of <math>z^k</math> is 1 if integer <math>k=-1</math> and 0 otherwise.

=== Generalization to Laurent series ===
If a function is expressed as a [[Laurent series]] expansion around c as follows:<math display="block">f(z) = \sum_{n=-\infty}^\infty a_n(z-c)^n.</math>Then, the residue at the point c is calculated as:<math display="block">\operatorname{Res}(f,c) = {1 \over 2\pi i} \oint_\gamma f(z)\,dz = {1 \over 2\pi i} \sum_{n=-\infty}^\infty \oint_\gamma a_n(z-c)^n \,dz = a_{-1} </math>using the results from contour integral of a monomial for counter clockwise contour integral <math>\gamma</math> around a point c. Hence, if a [[Laurent series]] representation of a function exists around c, then its residue around c is known by the coefficient of the <math>(z-c)^{-1}</math> term.

=== Application in residue theorem ===
{{Main|Residue theorem}}
For a [[meromorphic function]] <math>f</math>, with a finite set of singularities within a [[Curve orientation|positively oriented]] [[Jordan curve|simple closed curve]] <math>C</math> which does not pass through any singularity, the value of the contour integral is given according to [[residue theorem]], as:<math display="block">
\oint_C f(z)\, dz = 2\pi i \sum_{k=1}^n \operatorname{I}(C, a_k) \operatorname{Res}(f, a_k).
</math>where <math>\operatorname{I}(C, a_k)</math>, the winding number, is <math>1</math> if <math>a_k</math> is in the interior of <math>C</math> and <math>0</math> if not, simplifying to:<math display="block">
\oint_\gamma f(z)\, dz = 2\pi i \sum \operatorname{Res}(f, a_k)
</math>where <math>a_k</math> are all isolated singularities within the contour <math>C</math>.

== Calculation of residues ==
Suppose a [[punctured disk]] ''D'' = {''z'' : 0 < |''z'' &minus; ''c''| < ''R''} in the complex plane is given and ''f'' is a [[holomorphic function]] defined (at least) on ''D''. The residue Res(''f'', ''c'') of ''f'' at ''c'' is the coefficient ''a''<sub>&minus;1</sub> of {{nowrap|(''z'' &minus; ''c'')<sup>&minus;1</sup>}} in the [[Laurent series]] expansion of ''f'' around ''c''. Various methods exist for calculating this value, and the choice of which method to use depends on the function in question, and on the nature of the singularity.

According to the [[residue theorem]], we have:

: <math>\operatorname{Res}(f,c) = {1 \over 2\pi i} \oint_\gamma f(z)\,dz</math>

where ''γ'' traces out a circle around ''c'' in a counterclockwise manner and does not pass through or contain other singularities within it. We may choose the path ''γ'' to be a circle of radius ''ε'' around ''c.'' Since ''ε'' can be as small as we desire it can be made to contain only the singularity of c due to nature of isolated singularities. This may be used for calculation in cases where the integral can be calculated directly, but it is usually the case that residues are used to simplify calculation of integrals, and not the other way around.

===Removable singularities===
If the function ''f'' can be [[Analytic continuation|continued]] to a [[holomorphic function]] on the whole disk <math>|y-c|<R</math>, then Res(''f'',&nbsp;''c'')&nbsp;=&nbsp;0. The converse is not generally true.

===Simple poles===
If ''c'' is a [[simple pole]] of ''f'', the residue of ''f'' is given by:

:<math>\operatorname{Res}(f,c)=\lim_{z\to c}(z-c)f(z).</math>

If that limit does not exist, then ''f'' instead has an essential singularity at ''c''. If the limit is 0, then ''f'' is either analytic at ''c'' or has a removable singularity there. If the limit is equal to infinity, then the order of the pole is higher than 1.

It may be that the function ''f'' can be expressed as a quotient of two functions, <math>f(z)=\frac{g(z)}{h(z)}</math>, where ''g'' and ''h'' are [[holomorphic function]]s in a [[Neighbourhood (mathematics)|neighbourhood]] of ''c'', with ''h(c)'' = 0 and&nbsp;''h'(c)''&nbsp;≠&nbsp;0. In such a case, [[L'Hôpital's rule]] can be used to simplify the above formula to:

: <math>
\begin{align}
\operatorname{Res}(f,c) & =\lim_{z\to c}(z-c)f(z) = \lim_{z\to c}\frac{z g(z) - cg(z)}{h(z)} \\[4pt]
& = \lim_{z\to c}\frac{g(z) + z g'(z) - cg'(z)}{h'(z)} = \frac{g(c)}{h'(c)}.
\end{align}
</math>

===Limit formula for higher-order poles===
More generally, if ''c'' is a [[pole (complex analysis)|pole]] of order ''p'', then the residue of ''f'' around ''z'' = ''c'' can be found by the formula:

: <math> \operatorname{Res}(f,c) = \frac{1}{(p-1)!} \lim_{z \to c} \frac{d^{p-1}}{dz^{p-1}} \left( (z-c)^p f(z) \right). </math>

This formula can be very useful in determining the residues for low-order poles. For higher-order poles, the calculations can become unmanageable, and [[series expansion]] is usually easier. For [[essential singularity|essential singularities]], no such simple formula exists, and residues must usually be taken directly from series expansions.

===Residue at infinity===
In general, the [[residue at infinity]] is defined as:

: <math> \operatorname{Res}(f(z), \infty) = -\operatorname{Res}\left(\frac{1}{z^2} f\left(\frac 1 z \right), 0\right).</math>

If the following condition is met:

:<math> \lim_{|z| \to \infty} f(z) = 0,</math>

then the [[residue at infinity]] can be computed using the following formula:

:<math> \operatorname{Res}(f, \infty) = -\lim_{|z| \to \infty} z \cdot f(z).</math>

If instead

:<math> \lim_{|z| \to \infty} f(z) = c \neq 0,</math>

then the [[residue at infinity]] is

:<math> \operatorname{Res}(f, \infty) = \lim_{|z| \to \infty} z^2 \cdot f'(z).</math>
:

For functions meromorphic on the entire complex plane with finitely many singularities, the sum of the residues at the (necessarily) isolated singularities plus the residue at infinity is zero, which gives:

: <math> \operatorname{Res}(f(z), \infty) = -\sum_k \operatorname{Res} (f(z), a_k).</math>

=== Series methods ===
If parts or all of a function can be expanded into a [[Taylor series]] or [[Laurent series]], which may be possible if the parts or the whole of the function has a standard series expansion, then calculating the residue is significantly simpler than by other methods. The residue of the function is simply given by the coefficient of <math>(z-c)^{-1}</math> in the [[Laurent series]] expansion of the function.

== Examples<!-- Needs simple pole and higher order pole residue examples below --> ==

=== Residue from series expansion ===

==== Example 1 ====
As an example, consider the [[contour integral]]
:<math>\oint_C {e^z \over z^5}\,dz</math>
where ''C'' is some [[simple closed curve]] about 0.

Let us evaluate this integral using a standard convergence result about integration by series. We can substitute the [[Taylor series]] for <math>e^z</math> into the integrand. The integral then becomes

:<math>\oint_C {1 \over z^5}\left(1+z+{z^2 \over 2!} + {z^3\over 3!} + {z^4 \over 4!} + {z^5 \over 5!} + {z^6 \over 6!} + \cdots\right)\,dz.</math>

Let us bring the 1/''z''<sup>5</sup> factor into the series. The contour integral of the series then writes

: <math>
\begin{align}
& \oint_C \left({1 \over z^5}+{z \over z^5}+{z^2 \over 2!\;z^5} + {z^3\over 3!\;z^5} + {z^4 \over 4!\;z^5} + {z^5 \over 5!\;z^5} + {z^6 \over 6!\;z^5} + \cdots\right)\,dz \\[4pt]
= {} & \oint_C \left({1 \over\;z^5}+{1 \over\;z^4}+{1 \over 2!\;z^3} + {1\over 3!\;z^2} + {1 \over 4!\;z} + {1\over\;5!} + {z \over 6!} + \cdots\right)\,dz.
\end{align}
</math>

Since the series converges uniformly on the support of the integration path, we are allowed to exchange integration and summation.  
The series of the path integrals then collapses to a much simpler form because of the previous computation. So now the integral around ''C'' of every other term not in the form ''cz''<sup>&minus;1</sup> is zero, and the integral is reduced to

: <math>\oint_C {1 \over 4!\;z} \,dz= {1 \over 4!} \oint_C{1 \over z}\,dz={1 \over 4!}(2\pi i) = {\pi i \over 12}.</math>

The value 1/4! is the ''residue'' of ''e''<sup>''z''</sup>/''z''<sup>5</sup> at ''z'' = 0, and is denoted

: <math>\operatorname{Res}_0 {e^z \over z^5}, \text{ or } \operatorname{Res}_{z=0} {e^z \over z^5}, \text{ or } \operatorname{Res}(f,0) \text{ for } f={e^z \over z^5}.</math>

==== Example 2 ====
As a second example, consider calculating the residues at the singularities of the function<math display="block">f(z) = {\sin z \over z^2-z}</math>which may be used to calculate certain contour integrals. This function appears to have a singularity at ''z'' = 0, but if one factorizes the denominator and thus writes the function as<math display="block">f(z) = {\sin z \over z(z - 1)}</math>it is apparent that the singularity at ''z'' = 0 is a [[removable singularity]] and then the residue at ''z'' = 0 is therefore 0.
The only other singularity is at ''z'' = 1. Recall the expression for the Taylor series for a function ''g''(''z'') about ''z'' = ''a'':<math display="block"> g(z) = g(a) + g'(a)(z-a) + {g''(a)(z-a)^2 \over 2!} + {g'''(a)(z-a)^3 \over 3!}+ \cdots</math>So, for ''g''(''z'') = sin&nbsp;''z'' and ''a'' = 1 we have<math display="block"> \sin z = \sin 1 + (\cos 1)(z-1)+{-(\sin 1)(z-1)^2 \over 2!} + {-(\cos 1)(z-1)^3 \over 3!} + \cdots.</math>and for ''g''(''z'') = 1/''z'' and ''a'' = 1 we have<math display="block"> \frac{1}{z} = \frac1 {(z - 1) + 1} = 1 - (z - 1) + (z - 1)^2 - (z - 1)^3 + \cdots.</math>Multiplying those two series and introducing 1/(''z''&nbsp;−&nbsp;1) gives us<math display="block"> \frac{\sin z} {z(z - 1)} = {\sin 1 \over z-1} + (\cos 1 - \sin 1) + (z-1) \left(-\frac{\sin 1}{2!} - \cos1 + \sin 1\right) + \cdots.</math>So the residue of ''f''(''z'') at ''z'' = 1 is sin&nbsp;1.

==== Example 3 ====

The next example shows that, computing a residue by series expansion, a major role is played by the [[Formal series#The Lagrange inversion formula|Lagrange inversion theorem]]. Let<math display="block"> u(z) := \sum_{k\geq 1}u_k z^k</math>be an [[entire function]], and let<math display="block">v(z) := \sum_{k\geq 1}v_k z^k</math>with positive radius of convergence, and with <math display="inline"> v_1 \neq 0</math>. So <math display="inline"> v(z)</math> has a local inverse <math display="inline"> V(z)</math> at 0, and <math display="inline"> u(1/V(z))</math> is [[meromorphic]] at 0. Then we have:<math display="block">\operatorname{Res}_0 \big(u(1/V(z))\big) = \sum_{k=0}^\infty ku_k v_k. </math>Indeed,<math display="block">\operatorname{Res}_0\big(u(1/V(z))\big) = \operatorname{Res}_0 \left(\sum_{k\geq 1} u_k V(z)^{-k}\right) = \sum_{k\geq 1} u_k \operatorname{Res}_0 \big(V(z)^{-k}\big)</math>because the first series converges uniformly on any small circle around 0. Using the Lagrange  inversion theorem<math display="block">\operatorname{Res}_0 \big(V(z)^{-k}\big) = kv_k,</math>and we get the above expression. For example, if <math>u(z) = z + z^2</math> and also <math>v(z) = z + z^2</math>, then<math display="block">V(z) = \frac{2z}{1 + \sqrt{1 + 4z}}</math>and<math display="block">u(1/V(z)) = \frac{1 + \sqrt{1 + 4z}}{2z} + \frac{1 + 2z + \sqrt{1 + 4z}}{2z^2}.</math>The first term contributes 1 to the residue, and the second term contributes 2 since it is asymptotic to <math>1/z^2 + 2/z</math>.


Note that, with the corresponding stronger symmetric assumptions on <math display="inline"> u(z)</math> and <math display="inline"> v(z)</math>, it also follows<math display="block">\operatorname{Res}_0 \left(u(1/V)\right) = \operatorname{Res}_0\left(v(1/U)\right),</math>where <math display="inline"> U(z)</math> is a local inverse of <math display="inline"> u(z)</math> at&nbsp;0.

==See also==
* The [[residue theorem]] relates a contour integral around some of a function's poles to the sum of their residues
* [[Cauchy's integral formula]]
* [[Cauchy's integral theorem]]
* [[Mittag-Leffler's theorem]]
* [[Methods of contour integration]]
* [[Morera's theorem]]
* [[Partial fractions in complex analysis]]

==References==
*{{cite book|authorlink=Lars Ahlfors|first = Lars|last = Ahlfors|title = Complex Analysis|publisher = McGraw Hill|year = 1979}}
*{{cite book|last1=Marsden|first1=Jerrold E.|last2=Hoffman|first2=Michael J.|title=Basic Complex Analysis|publisher=W. H. Freeman|edition=3rd|isbn=978-0-7167-2877-1|year=1998|url=https://books.google.com/books?id=Z26tKIymJjMC&q=residue}}

== External links ==
* {{springer|title=Residue of an analytic function|id=p/r081560}}
* {{MathWorld | urlname= ComplexResidue | title= Complex Residue}}

[[Category:Meromorphic functions]]
[[Category:Complex analysis]]