TOPICS
Search

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, Pole, Residue Theorem

Portions of this entry contributed by Todd Rowland

Explore with Wolfram|Alpha

Cite this as:

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

Subject classifications