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Complex Residue


The constant a_(-1) in the Laurent series

 f(z)=sum_(n=-infty)^inftya_n(z-z_0)^n
(1)

of f(z) about a point z_0 is called the residue of f(z). If f is analytic at z_0, its residue is zero, but the converse is not always true (for example, 1/z^2 has residue of 0 at z=0 but is not analytic at z=0). The residue of a function f at a point z_0 may be denoted Res_(z=z_0)(f(z)). The residue is implemented in the Wolfram Language as Residue[f, {z, z0}].

Two basic examples of residues are given by Res_(z=0)1/z=1 and Res_(z=0)1/z^n=0 for n>1.

Residue

The residue of a function f around a point z_0 is also defined by

 Res_(z_0)f=1/(2pii)∮_gammafdz,
(2)

where gamma is counterclockwise simple closed contour, small enough to avoid any other poles of f. In fact, any counterclockwise path with contour winding number 1 which does not contain any other poles gives the same result by the Cauchy integral formula. The above diagram shows a suitable contour for which to define the residue of function, where the poles are indicated as black dots.

It is more natural to consider the residue of a meromorphic one-form because it is independent of the choice of coordinate. On a Riemann surface, the residue is defined for a meromorphic one-form alpha at a point p by writing alpha=fdz in a coordinate z around p. Then

 Res_(p)alpha=Res_(z=p)f.
(3)

The sum of the residues of intfdz is zero on the Riemann sphere. More generally, the sum of the residues of a meromorphic one-form on a compact Riemann surface must be zero.

The residues of a function f(z) may be found without explicitly expanding into a Laurent series as follows. If f(z) has a pole of order m at z_0, then a_n=0 for n<-m and a_(-m)!=0. Therefore,

f(z)=sum_(n=-m)^(infty)a_n(z-z_0)^n
(4)
=sum_(n=0)^(infty)a_(-m+n)(z-z_0)^(-m+n).
(5)

Multiplying both sides by (z-z_0)^m gives

 (z-z_0)^mf(z)=sum_(n=0)^inftya_(-m+n)(z-z_0)^n.
(6)

Take the first derivative and reindex,

d/(dz)[(z-z_0)^mf(z)]=sum_(n=0)^(infty)na_(-m+n)(z-z_0)^(n-1)
(7)
=sum_(n=1)^(infty)na_(-m+n)(z-z_0)^(n-1)
(8)
=sum_(n=0)^(infty)(n+1)a_(-m+n+1)(z-z_0)^n.
(9)

Take the second derivative and reindex,

(d^2)/(dz^2)[(z-z_0)^mf(z)]=sum_(n=0)^(infty)n(n+1)a_(-m+n+1)(z-z_0)^(n-1)
(10)
=sum_(n=1)^(infty)n(n+1)a_(-m+n+1)(z-z_0)^(n-1)
(11)
=sum_(n=0)^(infty)(n+1)(n+2)a_(-m+n+2)(z-z_0)^n.
(12)

Iterating then gives

(d^(m-1))/(dz^(m-1))[(z-z_0)^mf(z)]=sum_(n=0)^(infty)(n+1)(n+2)...(n+m-1)a_(n-1)(z-z_0)^n
(13)
=(m-1)!a_(-1)+sum_(n=1)^(infty)(n+1)(n+2)...(n+m-1)a_(n-1)(z-z_0)^n.
(14)

So

lim_(z->z_0)(d^(m-1))/(dz^(m-1))[(z-z_0)^mf(z)]=lim_(z->z_0)(m-1)!a_(-1)+0
(15)
=(m-1)!a_(-1)
(16)

since lim_(z->z_0)(z-z_0)^n=0, and the residue is

 a_(-1)=1/((m-1)!)(d^(m-1))/(dz^(m-1))[(z-z_0)^mf(z)]_(z=z_0).
(17)

The residues of a holomorphic function at its poles characterize a great deal of the structure of a function, appearing for example in the amazing residue theorem of contour integration.


See also

Cauchy Integral Formula, Cauchy Integral Theorem, Contour Integration, Contour Winding Number, Laurent Series, Meromorphic One-Form, Pole, Residue Theorem

Portions of this entry contributed by Todd Rowland

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Cite this as:

Rowland, Todd and Weisstein, Eric W. "Complex Residue." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/ComplexResidue.html

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