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| In [[mathematics]], more specifically [[complex analysis]], the '''residue''' is a [[complex number]] proportional to the [[line integral|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: \mathbb{C} \setminus \{a_k\} \rightarrow \mathbb{C}</math> that is [[holomorphic]] except at the discrete points {''a''<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]].
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| == Definition ==
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| The residue of a [[meromorphic function]] <math>f</math> at an [[isolated singularity]] <math>a</math>, often denoted <math>\operatorname{Res}(f,a)</math> or <math>\operatorname{Res}_a(f)</math>, is the unique value <math>R</math> such that <math>f(z)- R/(z-a)</math> has an [[Analytic function|analytic]] [[antiderivative]] in a [[punctured disk]] <math>0<\vert z-a\vert<\delta</math>.
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| 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]].
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| The definition of a residue can be generalized to arbitrary [[Riemann surfaces]]. Suppose <math>\omega</math> is a 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>
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| is defined to be the residue of <math>f(z)</math> at the point corresponding to <math>x</math>.
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| == Example ==
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| As an example, consider the [[contour integral]]
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| :<math>\oint_C {e^z \over z^5}\,dz</math>
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| where ''C'' is some [[simple closed curve]] about 0.
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| Let us evaluate this integral using a standard convergence result about integration by series. We can substitute the [[Taylor series]] for
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| <math>e^z</math> into the integrand. The integral then becomes
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| :<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>
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| Let us bring the 1/''z''<sup>5</sup> factor into the series. The contour integral of the series then writes
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| :: <math>\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 </math>
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| : <math>=\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.</math>
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| Since the series converges uniformly on the support of the integration path, we are allowed to exchange integration and summation.
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| The series of the path integrals then collapses to a much simpler form: recall that
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| :<math>\oint_C {1 \over z^n} \,dz=0,\quad n \in \mathbb{Z},\mbox{ for }n \ne 1.</math>
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| So now the integral around ''C'' of every other term not in the form ''cz''<sup>−1</sup> is zero, and the integral is reduced to
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| :<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>
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| The value 1/4! is the ''residue'' of ''e''<sup>''z''</sup>/''z''<sup>5</sup> at ''z'' = 0, and is denoted
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| :<math>\mathrm{Res}_0 {e^z \over z^5},\ \mathrm{or}\ \mathrm{Res}_{z=0} {e^z \over z^5},\ \mathrm{or}\ \mathrm{Res}(f,0).</math>
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| == Calculating residues ==
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| Suppose a [[punctured disk]] ''D'' = {''z'' : 0 < |''z'' − ''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>−1</sub> of (''z'' − ''c'')<sup>−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.
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| According to [[Cauchy's integral formula]], we have:
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| :<math>\operatorname{Res}(f,c) =
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| {1 \over 2\pi i} \oint_\gamma f(z)\,dz</math>
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| where ''γ'' traces out a circle around ''c'' in a counterclockwise manner. We may choose the path ''γ'' to be a circle of radius ''ε'' around ''c'', where ''ε'' is as small as we desire. 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.
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| ===Removable singularities===
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| If the function ''f'' can be [[Analytic continuation|continued]] to a holomorphic function on the whole disk { ''y'' :; ''c''| < ''R'' }, then Res(''f'', ''c'') = 0. The converse is not generally true.
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| ===Simple poles===
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| At a [[simple pole]] ''c'', the residue of ''f'' is given by:
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| :<math>\operatorname{Res}(f,c)=\lim_{z\to c}(z-c)f(z).</math>
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| It may be that the function ''f'' can be expressed as a quotient of two functions, ''f''(''z'')=''g''(''z'')/''h''(''z''), where ''g'' and ''h'' are [[holomorphic]] functions in a [[neighbourhood]] of ''c'', with ''h''(''c'') = 0 and ''h'''(''c'') ≠ 0. In such a case, the above formula simplifies to:
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| :<math>\operatorname{Res}(f,c) = \frac{g(c)}{h'(c)}.</math>
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| ===Limit formula for higher order poles===
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| More generally, if ''c'' is a [[pole (complex analysis)|pole]] of order ''n'', then the residue of ''f'' around ''z'' = ''c'' can be found by the formula:
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| :<math> \mathrm{Res}(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}}\left( (z-c)^{n}f(z) \right). </math>
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| 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. Also for [[essential singularity|essential singularities]], residues often must be taken directly from series expansions.
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| ===Residue at infinity===
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| If the following condition is met:
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| :<math> \lim_{|z| \to \infty} f(z) = 0</math>,
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| then the [[residue at infinity]] can be computed using the following formula:
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| :<math> \mathrm{Res}(f, \infty) = -\lim_{|z| \to \infty} z \cdot f(z)</math>.
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| If instead
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| :<math> \lim_{|z| \to \infty} f(z) = c \neq 0</math>,
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| then the [[residue at infinity]] is
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| :<math> \mathrm{Res}(f, \infty) = -\lim_{|z| \to \infty} z^2 \cdot f'(z)</math>.
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| The [[residue at infinity]] is also given by:
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| :<math> \mathrm{Res}(f(z), \infty) = -\mathrm{Res}\left(\frac{1}{z^2}f\left(\frac{1}{z}\right), 0\right)</math>.
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| === Series methods ===
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| 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.
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| '''1'''. As a first example, consider calculating the residues at the singularities of the function
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| : <math>f(z)={\sin{z} \over z^2-z}</math>
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| 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
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| : <math>f(z)={\sin{z} \over z(z-1)}</math>
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| it is apparent that the singularity at ''z'' = 0 is a [[removable singularity]] and then the residue at ''z'' = 0 is therefore 0. | |
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| The only other singularity is at ''z'' = 1. Recall the expression for the Taylor series for a function ''g''(''z'') about ''z'' = ''a'':
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| : <math> g(z) = g(a) + g'(a)(z-a) + {g''(a)(z-a)^2 \over 2!} + {g'''(a)(z-a)^3 \over 3!}+ \cdots</math>
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| So, for ''g''(''z'') = sin ''z'' and ''a'' = 1 we have
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| : <math> \sin{z} = \sin{1} + \cos{1}(z-1)+{-\sin{1}(z-1)^2 \over 2!} + {-\cos{1}(z-1)^3 \over 3!}+\cdots.</math>
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| and for ''g''(''z'') = 1/''z'' and ''a'' = 1 we have
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| : <math> \frac1z = \frac1 {(z-1)+1} = 1 - (z-1) + (z-1)^2 - (z-1)^3 + \cdots.</math>
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| Multiplying those two series and introducing 1/(''z'' − 1) gives us
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| : <math> \frac{\sin{z}} {z(z-1)} = {\sin{1} \over z-1} + (\cos{1}-\sin1) + (z-1) \left(-\frac{\sin{1}}{2!} - \cos1 + \sin1\right) + \cdots.</math>
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| So the residue of ''f''(''z'') at ''z'' = 1 is sin 1.
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| '''2'''. 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
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| :<math> u(z):=\sum_{k\geq 1}u_k z^k</math>
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| be an [[entire function]], and let
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| :<math>v(z):=\sum_{k\geq 1}v_k z^k</math>
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| with positive radius of convergence, and with <math>\textstyle v_1\neq 0 </math>. So <math>\textstyle v(z)</math> has a local inverse <math>\textstyle V(z)</math> at 0, and <math>\textstyle u(1/V(z))</math> is [[meromorphic]] at 0. Then we have:
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| :<math>\mathrm{Res_0}\big(u(1/V(z))\big)= \sum_{k=0}^{\infty} ku_k v_k</math>.
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| Indeed,
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| :<math>\mathrm{Res_0}\big(u(1/V(z))\big)=\mathrm{Res_0}\Big(\sum_{k\geq 1}u_k V(z)^{-k}\Big)=\sum_{k\geq 1} u_k \mathrm{Res_0}\big(V(z)^{-k}\big)
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| </math>
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| because the first series converges uniformly on any small circle around 0. Using the Lagrange inversion theorem
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| :<math>\mathrm{Res_0}\big(V(z)^{-k}\big)=kv_k</math>,
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| 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>V(z)=\frac{2z}{1+\sqrt{1+4z}}</math> and <math>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>.
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| Note that, with the corresponding stronger symmetric assumptions on <math>\textstyle u(z)</math> and <math>\textstyle v(z)</math>, it also follows
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| :<math>\mathrm{Res_0}\big(u(1/V)\big)=\mathrm{Res_0}\big(v(1/U)\big)</math> , | |
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| where <math>\textstyle U(z)</math> is a local inverse of <math>\textstyle u(z)</math> at 0.
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| ==See also==
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| * [[Cauchy's integral formula]]
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| * [[Cauchy's integral theorem]]
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| * [[Mittag-Leffler's theorem]]
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| * [[Methods of contour integration]]
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| * [[Morera's theorem]]
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| * [[Partial fractions in complex analysis]]
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| ==References==
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| *{{cite book|authorlink=Lars Ahlfors|first = Lars|last = Ahlfors|title = Complex Analysis|publisher = McGraw Hill|year = 1979}}
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| *{{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}}
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| == External links ==
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| * {{springer|title=Residue of an analytic function|id=p/r081560}}
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| * {{MathWorld | urlname= ComplexResidue | title= Complex Residue}}
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| * John H. Mathews. [http://math.fullerton.edu/mathews/c2003/ResidueCalcMod.html Module for Residues].
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| [[Category:Meromorphic functions]]
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