Residues and their applications

Residues
Calculus with Singular Points

Residues and their applications

First Order Pole

M-Order Pole

Residue Theorem

Residue at the Point at Infinity

Sum of Residues Theorem

Cauchy Integral Formula Generalization

Definite Integrals Evaluation

Trigonometric Functions

ANCHORTITLE

Let a point z_0 be an isolated singularity of a single-valued function f(z). In a punctured neighborhood of this point, f(z) admits a unique expansion into a Laurent series:

f(z) = \sum_{n=-\infty}^\infty c_n \left(z - z_0 \right)^n

The coefficients c_n are determined by the integral formula:

c_n = \frac{1}{2\pi i} \oint_{C} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} \mathrm d\zeta, \quad n \in \mathbb{Z}

where C is any simple closed contour in the annulus of convergence, oriented counter-clockwise, that encloses z_0. A coefficient of particular significance is c_{-1}, which is defined as the residue of the function f(z) at the singular point z_0.

The residue is formally defined by the integral:

\operatorname{Res} [f(z), z_0] = \frac{1}{2\pi i} \oint_C f(\zeta)\mathrm d\zeta

To see that this integral is indeed equal to c_{-1}, we can integrate the Laurent series term by term. The uniform convergence of the series on the compact contour C justifies the interchange of summation and integration:

\begin{aligned} \oint_C f(\zeta)\mathrm d\zeta &= \oint_C \sum_{n=-\infty}^\infty c_n \left(\zeta - z_0 \right)^n \mathrm d\zeta \\ &= \sum_{n=-\infty}^\infty c_n \oint_C \left(\zeta - z_0 \right)^n \mathrm d\zeta \end{aligned}

From the theory of complex integration, the contour integral of (\zeta - z_0)^n is zero for all integer values of n except for n=-1, for which its value is 2\pi i. Consequently, every term in the series vanishes except for the one corresponding to n=-1:

\oint_C f(\zeta)\mathrm d\zeta = c_{-1} (2\pi i)

This confirms the identity \operatorname{Res} [f(z), z_0] = c_{-1}. If z_0 is a regular point or a removable singularity, the principal part of the Laurent series is zero, meaning c_n=0 for all n < 0. In this situation, the residue is zero. The residue is a non-zero quantity only when z_0 is a pole or an essential singularity. The calculation of residues is a central operation in complex analysis, as it allows the evaluation of complex integrals to be reduced to an algebraic procedure of computing derivatives.

First order pole

Suppose z_0 is a simple pole (a pole of the first order) of the function f(z). The Laurent series expansion in the vicinity of this point has the form:

f(z) = c_{-1} (z - z_0)^{-1} + \sum_{n=0}^\infty c_n (z-z_0)^n

To isolate the residue c_{-1}, we can multiply the series by the factor (z - z_0):

(z - z_0) f(z) = c_{-1} + \sum_{n=0}^\infty c_n (z-z_0)^{n+1}

Taking the limit as z approaches z_0, all terms in the summation vanish, yielding a direct formula for the residue:

\operatorname{Res} [f(z), z_0] = c_{-1} = \lim_{z \to z_0} (z - z_0) f(z)

A frequent scenario is when f(z) is expressed as a ratio of two analytic functions, f(z) = g(z)/h(z), where g(z_0) \neq 0 and z_0 is a simple zero of h(z), meaning h(z_0) = 0 and h^\prime(z_0) \neq 0. In this case, the calculation of the residue simplifies further. Using the limit formula:

\begin{aligned} \operatorname{Res} [f(z), z_0] &= \lim_{z \to z_0} (z - z_0) \frac{g(z)}{h(z)} \\ &= \lim_{z \to z_0} g(z) \cdot \lim_{z \to z_0} \frac{z - z_0}{h(z) - h(z_0)} \\ &= g(z_0) \cdot \left[ \lim_{z \to z_0} \frac{h(z) - h(z_0)}{z - z_0} \right]^{-1} \end{aligned}

The second limit is the reciprocal of the definition of the derivative of h(z) at z_0. This gives the formula:

\operatorname{Res} \left[\frac{g(z)}{h(z)}, z_0\right] = \frac{g(z_0)}{h^\prime(z_0)}

M-order pole

Now, let z_0 be a pole of order m > 1 for the function f(z). The Laurent series expansion is:

f(z) = \frac{c_{-m}}{(z - z_0)^m} + \dots + \frac{c_{-1}}{z - z_0} + \sum_{n=0}^\infty c_n (z-z_0)^n

To find c_{-1}, we first multiply f(z) by (z - z_0)^m. This operation removes the singularity at z_0 and results in a function, let’s call it g(z), which is analytic at z_0.

g(z) = (z - z_0)^m f(z) = c_{-m} + c_{-m+1}(z-z_0) + \dots + c_{-1}(z-z_0)^{m-1} + \sum_{n=0}^\infty c_n (z-z_0)^{n+m}

The expression on the right is the Taylor series expansion of the analytic function g(z) around the point z_0. The coefficient c_{-1} is the coefficient of the (z-z_0)^{m-1} term in this Taylor series. The standard formula for the coefficients of a Taylor series states that the coefficient of (z-z_0)^k for a function g(z) is given by \frac{1}{k!} \frac{\mathrm d^k g}{\mathrm d z^k}\bigg|_{z=z_0}.

Applying this formula for k=m-1, we identify c_{-1}:

c_{-1} = \frac{1}{(m-1)!} \frac{\mathrm d^{m-1}g(z)}{\mathrm d z^{m-1}}\bigg|_{z=z_0}

Substituting the expression for g(z) back yields the general formula for the residue at a pole of order m:

\operatorname{Res} [f(z), z_0] = \frac{1}{(m-1)!}\lim_{z \to z_0} \frac{\mathrm d^{m-1}}{\mathrm d z^{m-1}} \left[ (z - z_0)^m f(z)\right]

For the case of a simple pole (m=1), it reduces to the previous result.

Residue theorem

Theorem: Let a function f(z) be analytic inside and on a simple closed contour C, except for a finite number of isolated singular points z_1, z_2, \dots, z_n located inside C. Then, the integral of f(z) taken counter-clockwise along C is proportional to the sum of the residues of the function at these singular points:

\oint_C f(\zeta) \, \mathrm d \zeta = 2\pi i \sum_{k=1}^{n} \operatorname{Res}[f(z), z_k]

The proof of this theorem can be found here. The utility of this result is that it transforms the often-difficult task of direct integration into the algebraic problem of calculating residues.

Residue at the point at infinity

The concept of a residue can be extended to the point at infinity. Let z=\infty be an isolated singular point for an analytic function f(z). This implies that f(z) is analytic in an annulus R < |z| < \infty for some sufficiently large R.

The residue of f(z) at the point z = \infty is defined by the integral:

\operatorname{Res}[f(z), \infty] = \frac{1}{2\pi i}\oint_{C^-} f(\zeta) \, \mathrm d \zeta = - \frac{1}{2\pi i}\oint_{C^+} f(\zeta) \, \mathrm d \zeta

The integration contour C is any simple closed curve, such as a circle |z| = \rho > R, outside of which f(z) has no finite singularities. The notation C^- signifies that the contour is traversed in the clockwise (negative) direction. This orientation is chosen so that the unbounded region containing z=\infty is kept to the left, analogous to how the interior is kept to the left for a standard counter-clockwise (C^+) contour.

To relate this definition to the Laurent series of f(z) at infinity, we perform a change of variables z = 1/\zeta. The clockwise path C^- for z becomes a counter-clockwise path \gamma^+ for \zeta.

\operatorname{Res}[f(z), \infty] = -\frac{1}{2\pi i} \oint_{C^+} f(z) \mathrm dz = -\frac{1}{2\pi i} \oint_{\gamma^+} f(1/\zeta) \left(-\frac{1}{\zeta^2}\right) \mathrm d\zeta

The Laurent series for f(1/\zeta) around \zeta=0 is:

f(1/\zeta) = \sum_{n=-\infty}^\infty c_n \zeta^{-n} = \dots + c_{-1}\zeta + c_0 + c_1\zeta^{-1} + \dots

The integral becomes:

\operatorname{Res}[f(z), \infty] = \frac{1}{2\pi i} \oint_{\gamma^+} \frac{1}{\zeta^2} \sum_{n=-\infty}^\infty c_n \zeta^{-n} \mathrm d\zeta = \frac{1}{2\pi i} \oint_{\gamma^+} \sum_{n=-\infty}^\infty c_n \zeta^{-n-2} \mathrm d\zeta

By the term-by-term integration of the series, the only non-zero contribution comes from the term where the power of \zeta is -1. This occurs when -n-2 = -1, which implies n=-1. The coefficient of this term is c_{-1}. Therefore, the integral evaluates to c_{-1}. The calculation shows:

\operatorname{Res}[f(z), \infty] = -c_{-1}

The negative sign is a direct result of the definition. This leads to a distinction: while the residue at a finite removable singularity is always zero (since c_{-1}=0), the residue at a removable singularity at infinity can be non-zero. For instance, for f(z)=1/z, the point z=\infty is a removable singularity. The Laurent series at infinity is f(z) = z^{-1}, so c_{-1}=1. The residue at infinity is \operatorname{Res}[f(z), \infty]=-c_{-1}=-1.

Sum of residues theorem

A consequence of this theory is that the sum of all residues of a function over the entire extended complex plane is zero.

Theorem: If a function f(z) is analytic on the extended complex plane except for a finite number of isolated singular points \{z_1, \dots, z_{N-1}\} and z_N = \infty, then the sum of all its residues is zero:

\sum_{k=1}^N \operatorname{Res} [f(z), z_k] = 0

Proof: Let C be a closed contour, oriented counter-clockwise, that encloses all the finite singular points z_1, \dots, z_{N-1}. According to the residue theorem:

\frac{1}{2\pi i}\oint_{C^+} f(\zeta) \, \mathrm d \zeta = \sum_{k=1}^{N-1} \operatorname{Res}[f(z), z_k]

From the definition of the residue at infinity, the same integral is related to \operatorname{Res}[f(z), \infty]:

\operatorname{Res} [f(z), \infty] = - \frac{1}{2\pi i}\oint_{C^+} f(\zeta) \, \mathrm d \zeta

Combining these two equations gives:

\sum_{k=1}^{N-1} \operatorname{Res}[f(z), z_k] = -\operatorname{Res} [f(z), \infty]

Rearranging the terms completes the proof.

Cauchy integral formula generalization

The sum of residues theorem provides a method to simplify the evaluation of certain contour integrals. Suppose we need to compute the integral of f(z) along a closed contour C, and there are numerous singularities inside C, making a direct application of the residue theorem laborious. If, however, there are only a few singularities outside C, we can use the sum theorem.

Let \{z_k^{\text{in}}\} be the singularities inside C and \{z_k^{\text{out}}\} be those outside. The theorem states:

\sum \operatorname{Res}[f, z_k^{\text{in}}] + \sum \operatorname{Res}[f, z_k^{\text{out}}] + \operatorname{Res}[f, \infty] = 0

Since \oint_{C^+} f(\zeta) \, \mathrm d \zeta = 2\pi i \sum \operatorname{Res}[f, z_k^{\text{in}}], we can write:

\oint_{C^+} f(\zeta) \, \mathrm d \zeta = -2\pi i \left( \textstyle \sum_{k} \operatorname{Res}[f, z_k^{\text{out}}] + \operatorname{Res} [f(z), \infty] \right)

This formula can be used in generalizing Cauchy’s integral formula to unbounded domains. Let f(z) be analytic everywhere outside a simple closed contour C. Assume z=\infty is a removable singularity of f(z), so that \lim_{z\to \infty} f(z) = f(\infty) exists and is finite. We wish to evaluate the integral of \frac{f(\zeta)}{\zeta-z_0} around C.

Consider the auxiliary function g(z) = \frac{f(z)}{z - z_0}. The singularities of g(z) outside of C can only be at z=z_0 and at z=\infty.

The residue of g(z) at infinity is:

\operatorname{Res}[g(z), \infty] = -c_{-1}^g

where c_{-1}^g is the coefficient of z^{-1} in the Laurent expansion of g(z) for large z. For large z:

\begin{aligned} g(z) &= \frac{1}{z-z_0}f(z) = \frac{1}{z(1-z_0/z)} \left(f(\infty) + \frac{c_{-1}}{z} + \dots \right) \\ &= \frac{1}{z}\left(1+\frac{z_0}{z}+\dots\right)\left(f(\infty) + \frac{c_{-1}}{z} + \dots \right) \\ &= \frac{f(\infty)}{z} + O(z^{-2}) \end{aligned}

The coefficient c_{-1}^g is f(\infty), so \operatorname{Res}[g(z), \infty] = -f(\infty).

Now we analyze two cases based on the position of z_0.

Case 1: z_0 is in the bounded domain interior to C. The function g(z) is analytic everywhere outside C. The only exterior singularity is at infinity. Applying our formula:

\oint_{C^+} \frac{f(\zeta)}{\zeta - z_0} \, \mathrm d \zeta = -2\pi i \operatorname{Res}[g(z), \infty] = -2\pi i(-f(\infty)) = 2\pi i f(\infty)

Case 2: z_0 is in the unbounded domain exterior to C. In this case, g(z) has two singularities outside C: a simple pole at z=z_0 and the point at infinity. The residue at the pole is:

\operatorname{Res}[g(z), z_0] = \lim_{z \to z_0} (z-z_0) \frac{f(z)}{z-z_0} = f(z_0)

The sum of the exterior residues is \operatorname{Res}[g(z), z_0] + \operatorname{Res}[g(z), \infty] = f(z_0) - f(\infty). The integral is:

\oint_{C^+} \frac{f(\zeta)}{\zeta - z_0} \, \mathrm d \zeta = -2\pi i \left( f(z_0) - f(\infty) \right) = 2\pi i \left( f(\infty) - f(z_0) \right)

Combining these results gives a generalized Cauchy integral formula for an exterior domain, where C is traversed counter-clockwise:

\frac{1}{2\pi i}\oint_{C^+} \frac{f(\zeta)}{\zeta - z_0} \, \mathrm d \zeta = \begin{cases} f(\infty) & \text{if } z_0 \text{ is inside } C \\ f(\infty) - f(z_0) & \text{if } z_0 \text{ is outside } C \end{cases}

Definite integrals evaluation

Certain classes of integrals, which can be challenging to compute with standard real analysis techniques, become quite manageable when transformed into contour integrals in the complex plane.

Trigonometric functions

A common class of definite integrals for which these methods can be used involves rational functions of trigonometric expressions, integrated over a full period.

Consider an integral of the general form:

I = \int_0^{2\pi} f(\cos(\theta), \sin(\theta)) \, \mathrm d\theta

where f is a rational function of its two arguments. The evaluation strategy involves a change of variables that maps this real integral into a contour integral over the unit circle in the complex plane.

Let the complex variable z be defined as z = e^{i \theta}. As the real variable \theta ranges from 0 to 2\pi, the complex variable z traverses the unit circle, |z|=1, in the counter-clockwise direction.

From this substitution, we derive the following expressions for the differential and the trigonometric functions:

\begin{aligned} &\mathrm dz = i e^{i \theta} \mathrm d\theta = i z \mathrm d\theta \\ &\mathrm d\theta = \frac{\mathrm dz}{iz} \end{aligned}

The trigonometric functions \cos(\theta) and \sin(\theta) can be expressed in terms of z using Euler’s formula:

\begin{aligned} & \cos(\theta) = \frac{1}{2}\left(e^{i \theta} + e^{-i \theta}\right) = \frac{1}{2}\left(z + \frac{1}{z}\right) \\ & \sin(\theta) = \frac{1}{2i}\left(e^{i \theta} - e^{-i \theta}\right) = \frac{1}{2i}\left(z - \frac{1}{z}\right) \end{aligned}

Substituting these expressions transforms the original real integral into a complex contour integral:

I = \frac{1}{i} \oint_{|z|=1} F\left(z + \frac{1}{z}, z - \frac{1}{z} \right) \frac{\mathrm dz}{z}

The integrand is now a new function, let’s call it G(z), which is a rational function of z.

Assuming G(z) is analytic within the contour |z|=1 except for a finite number of poles, the integral can be evaluated using the residue theorem:

I = 2\pi i \sum_{k} \operatorname{Res}[G(z), z_k]

where the sum is taken over all poles z_k of G(z) that are located inside the unit circle, |z_k| < 1. If a pole z_k is of order \beta_k, its residue is calculated by the general formula:

\operatorname{Res}[G(z), z_k] = \frac{1}{(\beta_k-1)!} \lim_{z \to z_k} \frac{\mathrm d^{\beta_k-1}}{\mathrm d z^{\beta_k-1}} \left[ (z - z_k)^{\beta_k} G(z) \right]

For example, let’s evaluate the integral:

I = \int_0^{2\pi} \frac{\mathrm d\theta}{1 + a \, \cos(\theta)}, \quad |a|< 1

Applying the substitution z = e^{i\theta}:

\begin{aligned} I &= \oint_{|z|=1} \frac{1}{1 + a \left(\frac{z + z^{-1}}{2}\right)} \frac{\mathrm dz}{iz} \\ &= \oint_{|z|=1} \frac{1}{1 + \frac{a}{2z}(z^2 + 1)} \frac{\mathrm dz}{iz} \\ &= \frac{1}{i} \oint_{|z|=1} \frac{2z}{2z + a(z^2 + 1)} \frac{\mathrm dz}{z} \\ &= \frac{2}{i} \oint_{|z|=1} \frac{1}{az^2 + 2z + a} \mathrm dz \end{aligned}

The integrand has singularities where the denominator is zero. The roots of the quadratic equation az^2 + 2z + a = 0 are:

z_{1,2} = \frac{-2 \pm \sqrt{4 - 4a^2}}{2a} = \frac{-1 \pm \sqrt{1 - a^2}}{a}

These are two simple poles. Let’s analyze their location. The product of the roots is z_1 z_2 = a/a = 1. Since |z_1||z_2| = 1 and the roots are distinct for |a|<1, one root must lie inside the unit circle and the other outside. Let us identify the root with modulus less than one:

z_1 = \frac{-1 + \sqrt{1 - a^2}}{a}

For |a|<1, the magnitude of z_1 is less than 1. The integral’s value is determined by the residue at this pole. Since it is a simple pole of the form P(z)/Q(z), the residue is P(z_1)/Q^\prime(z_1). Here, P(z) = 1 and Q(z) = az^2 + 2z + a, so Q^\prime(z) = 2az + 2.

Applying the residue theorem, we obtain the final result for the integral:

\begin{aligned} I &= \frac{2}{i} 2\pi i \operatorname{Res}\left[\frac{1}{az^2+ 2z + a}, z_1 \right] =4\pi \frac{1}{2az_1 + 2} \\ & = 4\pi\frac{1}{2a\left(\frac{-1 + \sqrt{1 - a^2}}{a}\right) + 2} \\ & = 4\pi\frac{1}{-2 + 2\sqrt{1-a^2} + 2} = 4\pi\frac{1}{2\sqrt{1-a^2}} \\ & = \frac{2\pi}{\sqrt{1-a^2}} \end{aligned}

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