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Residue (Complex Analysis)In complex analysis, the residue is a complex number which describes the behavior of path integrals of a meromorphic function around a singularity. Residues can be computed quite easily and, once known, allow the determination of more complicated path integrals via the residue theorem. Motivation As an example, consider the contour integral -
where C is some Jordan curve about 0. Let us evaluate this integral without using standard integral theorems that may be available to us. Now, the Taylor series for ez is well-known, and we substitute this series into the integrand. The integral then becomes: -
Let us bring the 1/z5 term into the series, and so, we obtain -
-
The integral now collapses to a much simpler form. Recall -
So now the integral around C of every other term not in the form cz−1 becomes zero, and the integral is reduced to -
The value 1/4! is known as the residue of ez/z5 at z=0, and is notated as -
Calculating residues 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−1 of (z − c)−1 in the Laurent series expansion of f around c. At a simple pole, the residue is given by: -
According to the integral formula given in the Laurent series article we have: -
{1 \over 2\pi i} \int_\gamma f(z)\,dz where γ traces out a circle around c in a counterclockwise manner. We may choose the path γ to be a circle in radius ε around c were ε is as small as we desire. To calculate the residue of a function around z = c, a pole of order n, one may use the following formula: -
If the function f can be continued to a holomorphic function on the whole disk { z : |z − c| < R }, then Res(f, c) = 0. The converse is not generally true. 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, calculating the residue is significantly simpler than by other methods. As an example, consider calculating the residues at the singularities of the function -
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 -
it is apparent that the singularity at z=0 is a removable singularity and thus the residue at z=0 is therefore 0. The only other singularity is at z=1. Recall -
about z=a, so, for g(z)=sin z and a=1 we have -
Introducing 1/(z-1) gives us -
So the residue of f(z) at z=1 is sin 1.
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