Cauchy's Residue Theorem
Theorem: Let \mathcal{D} be a simply connected open subset of the complex plane \mathbb{C}. Let f be a function that is analytic on \mathcal{D} except for a finite set of isolated singularities \{z_1, z_2, \dots, z_n\} contained within \mathcal{D}. Let C be a positively oriented, simple closed contour in \mathcal{D} that encloses all of these singularities and on which f(z) is analytic. The contour integral of f along C is then given by the sum of the residues of f at each enclosed singularity, multiplied by 2\pi i.
\oint_C f(z) \, \mathrm d z = 2\pi i \sum_{k=1}^{n} \operatorname{Res}(f, z_k)
Proof: we will use the deformation invariance theorem, which is a extension of Cauchy-Goursat’s integral theorem to multiply connected domains. The strategy is to deform the original contour C into a new composite contour that isolates each singularity.
Consider the function f(z) and the contour C as described in the theorem statement. For each isolated singularity z_k, we can construct a small circle, C_k, centered at z_k with a radius r_k small enough such that each circle is contained entirely within the interior of C and none of the circles intersect.
We now construct a new region where f(z) is analytic. This is achieved by creating a new composite contour. Imagine making a narrow cut from the main contour C to each small circle C_k.
We form a new path C^\prime which traverses C in the counter-clockwise direction, then moves along a cut to the first circle C_1, traverses C_1 in the clockwise direction, returns along the cut, and repeats this process for all singularities z_k.
The region enclosed by this new composite contour C^\prime is simply connected, and the function f(z) is analytic everywhere within and on this new contour. By Cauchy-Goursat’s integral theorem (here), the integral over this composite path must be zero:
\oint_{C^\prime} f(z) \, \mathrm d z = 0
The integral over C^\prime can be decomposed into the integral over the outer contour C, the integrals over the small circles C_k, and the integrals along the connecting cuts.
Each cut is traversed twice, once in each direction. Since the paths are identical but the orientation is opposite, the contributions from these cuts cancel each other out. Let C_k^{-} denote the clockwise traversal of the circle C_k.
The vanishing integral over C^\prime can be expressed as:
\oint_C f(z) \, \mathrm d z + \sum_{k=1}^{n} \oint_{C_k^{-}} f(z) \, \mathrm d z = 0
Reversing the orientation of integration for each C_k^{-} introduces a negative sign, changing the traversal to the standard positive (counter-clockwise) direction, denoted by C_k:
\oint_C f(z) \, \mathrm d z - \sum_{k=1}^{n} \oint_{C_k} f(z) \, \mathrm d z = 0
This leads to the relationship:
\oint_C f(z) \, \mathrm d z = \sum_{k=1}^{n} \oint_{C_k} f(z) \, \mathrm d z
The next step is to evaluate the integral around a single isolated singularity z_k.
We employ the Laurent series expansion of f(z) in a punctured disk centered at z_k, which converges uniformly on the compact contour C_k:
f(z) = \sum_{j=-\infty}^{\infty} a_j (z - z_k)^j = \sum_{j=0}^{\infty} a_j (z - z_k)^j + \sum_{j=1}^{\infty} a_{-j} (z - z_k)^{-j}
Substituting the series into the integral and exchanging the order of summation and integration, which is permissible due to uniform convergence, we get:
\oint_{C_k} f(z) \, \mathrm d z = \oint_{C_k} \sum_{j=-\infty}^{\infty} a_j (z - z_k)^j \, \mathrm d z = \sum_{j=-\infty}^{\infty} a_j \oint_{C_k} (z - z_k)^j \, \mathrm d z
To evaluate the term-wise integrals, we parametrize the circular contour C_k by z(t) = z_k + r_k e^{it} for t \in [0, 2\pi], which implies \mathrm d z = i r_k e^{it} \, \mathrm d t:
\begin{aligned} \oint_{C_k} (z - z_k)^j \, \mathrm d z &= \int_0^{2\pi} (r_k e^{it})^j (i r_k e^{it}) \, \mathrm d t \\ &= i r_k^{j+1} \int_0^{2\pi} e^{i(j+1)t} \, \mathrm d t \end{aligned}
This integral has two outcomes depending on the integer value of j.
For j \neq -1:
i r_k^{j+1} \left[ \frac{e^{i(j+1)t}}{i(j+1)} \right]_0^{2\pi} = \frac{r_k^{j+1}}{j+1} (e^{i2\pi(j+1)} - e^0) = \frac{r_k^{j+1}}{j+1} (1 - 1) = 0
For j = -1:
i r_k^{0} \int_0^{2\pi} e^{0 \cdot t} \, \mathrm d t = i \int_0^{2\pi} 1 \, \mathrm d t = 2\pi i
The only term from the Laurent series that yields a non-zero contribution to the integral is the term where j=-1. The coefficient of this term, a_{-1}, is by definition the residue of f(z) at z_k, denoted \operatorname{Res}(f, z_k).
Therefore, the integral around a single singularity is:
\oint_{C_k} f(z) \, \mathrm d z = a_{-1} (2\pi i) = 2\pi i \, \operatorname{Res}(f, z_k)
Substituting this result back into our expression for the integral along C, we obtain the final statement of the theorem:
\oint_C f(z) \, \mathrm d z = \sum_{k=1}^{n} (2\pi i \, \operatorname{Res}(f, z_k)) = 2\pi i \sum_{k=1}^{n} \operatorname{Res}(f, z_k)
This completes the proof.
Residue calculation formulas
The definition of the residue as the coefficient a_{-1} is correct, but direct computation of the Laurent series is not always practical.
Below, we derive formulas for calculating the residue for poles of finite order.
Simple pole
If z_0 is a simple pole (a pole of order m=1), the Laurent series of f(z) around z_0 takes the form:
f(z) = \frac{a_{-1}}{z-z_0} + \sum_{j=0}^{\infty} a_j (z - z_0)^j
The summation term represents the analytic part of the series.
To isolate the residue a_{-1}, we can multiply the entire expression by (z-z_0):
(z-z_0)f(z) = a_{-1} + \sum_{j=0}^{\infty} a_j (z - z_0)^{j+1}
Taking the limit as z \to z_0, every term in the summation vanishes.
\lim_{z \to z_0} (z-z_0)f(z) = a_{-1}
This provides the well-known formula for the residue at a simple pole:
\operatorname{Res}(f, z_0) = \lim_{z \to z_0} (z-z_0)f(z)
Pole of order m
If z_0 is a pole of order m, the Laurent series is:
f(z) = \frac{a_{-m}}{(z-z_0)^m} + \dots + \frac{a_{-1}}{z-z_0} + \sum_{j=0}^{\infty} a_j (z - z_0)^j
Our objective is to isolate a_{-1}.
We begin by multiplying by (z-z_0)^m, which removes the singular part and yields an analytic function, which we denote \phi(z):
\phi(z) = (z-z_0)^m f(z) = a_{-m} + a_{-(m-1)}(z-z_0) + \dots + a_{-1}(z-z_0)^{m-1} + \sum_{j=0}^{\infty} a_j (z - z_0)^{j+m}
The function \phi(z) is analytic at z_0, and its power series expansion around z_0 is given above.
The coefficient a_{-1} is the coefficient of the (z-z_0)^{m-1} term.
From the theory of Taylor series, we know that the coefficient of the (z-z_0)^k term in the expansion of an analytic function is given by \phi^{(k)}(z_0)/k!.
To extract a_{-1}, we must differentiate \phi(z) a total of (m-1) times. After performing these, all terms (z-z_0)^k with k < m-1 will vanish.
The term a_{-1}(z-z_0)^{m-1} will become the constant a_{-1}(m-1)! and dll higher-order terms will still contain a factor of (z-z_0):
\frac{\mathrm d^{m-1}}{\mathrm d z^{m-1}} \phi(z) = a_{-1}(m-1)! + m! a_0 (z-z_0) + \dots
Taking the limit as z \to z_0 eliminates all terms except for the constant term:
\lim_{z \to z_0} \frac{\mathrm d^{m-1}}{\mathrm d z^{m-1}} \phi(z) = a_{-1}(m-1)!
Solving for a_{-1} and substituting back \phi(z) = (z-z_0)^m f(z), we arrive at the general formula for the residue at a pole of order m:
\operatorname{Res}(f, 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]
This general formula encompasses the simple pole case, as setting m=1 recovers the previous result, noting that the zeroth derivative is the function itself and 0! = 1.