Calculus of Residues - Complex Analysis, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

The Residue Theorem

A very important application of the theory of analytic functions involves the evaluation of real denite integrals. The key ingredients in the evaluation procedure are the concept of a residue and an associated theorem. Thus, our rst task is to familiarize ourselves with both.

Denition: Let f(z) be holomorphic in some deleted neighbourhood of z = z₀ , 0 < |z − z₀ | < R say, and let C be any closed contour within this neighbourhood and enclosing z = z₀ . Then, the integral is independent of the choice of C and is called the residue of f (z) at z = z₀.

                (3.1.1)

Since f (z) is holomorphic in 0 < |z − z₀ | < R , it must possess the Laurent series representation

                              (3.1.2)

where C is any closed contour in the annulus 0 < |ζ − z₀| < R that encloses ζ = z₀. Comparing (3.1.1) and (3.1.2) we see that an equivalent denition of the residue of f (z) at z = z₀ is

Res[f (z₀)] = c−1 .                          (3.1.3)

Equation (3.1.3) applies only to points in the nite plane. However, our rst de-nition (3.1.1) can be applied at innity as well, provided one does so with care. If f (z) is holomorphic or has an isolated singularity at z = ∞, it must be possible to dene a large circle C that encloses all the nite singularities of f (z). The circle C lies in an annulus R < |z| < ∞ in which f (z) is holomorphic and it encloses the point at innity.
Thus, (3.1.1) may be used with this curve to dene

                        (3.1.4)

where the minus sign is due to an anticlockwise circuit with respect to z = ∞ being a clockwise circuit with respect to the origin. Let us now apply the transformation z = 1/ζ to the integral in (3.1.4). Since this transformation again reverses the sense of the integration, and  we obtain

                         (3.1.5)

where C0 is a small circle about ζ = 0. From Cauchy’s Integral we then have provided this limit exists.

                       (3.1.6)

This formula brings out an interesting distinction between residues at innity and residues at points in the nite plane. If f (z) is holomorphic at z = z₀, then Res[f (z₀)] = 0 by Cauchy’s Theorem. But if f (z) is holomorphic at innity, then in general Res[f (∞)] ≠ 0. For example, the rational function f (z) = c/z has Res[f (0)] = c and Res[f (∞)] = −c even though it is clearly holomorphic at the latter point. Cauchy’s Residue Theorem implies a relationship between a function’s residue at innity and its residues in the finite plane which in turn, determines whether the former will vanish.

Theorem: If f (z) is holomorphic on and within a closed contour C except for a nite number of isolated singularities at z = z₁ , z₂ , . . . , z_n inside C , then

                       (3.1.7)

Proof : The proof of this theorem involves little more than an application of the generalized Cauchy Theorem.

As we did for our last theorem, we individually enclose each singularity z = z_k with a small circle C_k contained within C . Then, since f (z) is holomorphic within and on the boundary of the (n + 1)-fold connected domain bounded by C , C₁ , . . . , C_n , we can apply equation (2.2.8) and write

Invoking the denition of a residue, equation (3.1.1), this immediately yields as required.

If a function’s singularities are all isolated, then it follows from (3.1.4) and (3.1.7) that where the sum is taken over all singular points in the nite plane. Thus, it is the function’s behaviour throughout the nite plane that determines whether its residue at innity will vanish.

An immediate application of the residue theorem is the evaluation of closed contour integrals. Provided that the integrand possesses only isolated singularities within (or alternatively, without) the contour, the evaluation is reduced to the considerably less arduous task of calculating residues.

Calculating Residues

The basis of all residue calculus techniques is equation (3.1.3) which identies the residue of a function f (z) at the point z = z₀ with the coeficient of the rst inverse power of (z − z₀) in the Laurent expansion of f (z) about z = z₀. The most direct approach and the only one available when z = z₀ is an essential singularity is to use one of the practical methods (see Section 2.4.4) available for generating power series and determine the one coeficient we need. However, if z = z0 is a pole of order n , there is an alternative approach which is frequently but by no means invariably more convenient. It is based on the fact that within the annulus of convergence, 0 < |z − z₀| < R , of the Laurent expansion of f (z) about z = z₀ , we may set

f (z) = (z − z₀)⁻ⁿ g(z)                       (3.2.1)

where g(z) is holomorphic within |z − z₀| < R and is non-zero at z = z₀ . Putting (3.2.1) into the dening equation (3.1.1), we have

where we have used Cauchy’s Dierentiation Formula (2.3.5) in the last step. Thus, substituting for g(z) we obtain

                        (3.2.2)

In the case of a simple pole (n = 1) this expression reduces to

                        (3.2.3)

and, in the case of a function of the form f (z)   g(z0 ) ≠ 0, (3.2.3) in turn reduces to

                   (3.2.4)

Simple poles arising from simple zeros are suffciently common that this last special case will become memorized through use.

Examples: Consider the rational function

By performing a partial fraction decomposition,

we can read o the values of the residues at z = 2 and z = −1. Since the second term is holomorphic at z = 2, there can be no term involving the power (z − 2)−1 in the Laurent expansion of f (z) about z = 2. Therefore, Res[f (2)] = 0. Similarly, since the rst term is holomorphic at z = −1, we have Res[f (−1)] = 1. These results can be veried by using equation (3.2.2) . For example,

Next, suppose that we wish to calculate the residue of

at z = 1 where it has a fourth-order pole. Use of (3.2.2) would involve the calculation of the third derivative of  which, while not overly diffcult, is suffciently lengthy to pose some risk of error. On the other hand, the Laurent series approach in this case is relatively straightforward and hence, less likely to give rise to lost minus signs or factors of two.

We need to expand both e^3z and (z + 2)⁻¹ about z = 1. The exponential is entire and has the mth derivative

Therefore,

The function (z + 2)⁻¹ is holomorphic in |z − 1| < 3 and so admits the Taylor series.

Now, to obtain the coeffcient of (z − 1)−1 in the Laurent expansion of    about z = 1, we need only determine the coeffcient of (z − 1)³ in the product series

Thus,

Suppose now that we wish to integrate

around the circle |z| = 2. Writing f (z) in the form

we see that it is holomorphic at z = 0 but possesses rst order poles at z = ±π /2 arising from the rst order zeros of cos z at these two points. We can use equation (3.2.4) to calculate the residues at the poles; we find

Without further eort, the Residue Theorem gives us the following value for the integral in question:

A somewhat more difficult and more interesting problem is posed by the integral

where C is the unit circle, |z| = 1. Both 1/z and cosec z have rst order poles at z = 0, but sin 1/z has an essential singularity there. Therefore, we have no choice in this case: the Laurent series method of calculating residues is the only one that is applicable.

We require the expansions of cosec z and sin 1/z about z = 0. These were found in Section 2.2.4 and are

Because of the presence of the 1z factor in the integrand, we now must nd the coefficient of (z)⁰ in the Cauchy product of these two series. This is easily accomplished and yields the convergent series

Therefore, our integral has the value

The most important application of the Residue Theorem is in the evaluation of certain types of real definite integrals.