Residue Theorem (original) (raw)

Last Updated : 14 Nov, 2025

The Residue Theorem is a powerful tool in complex analysis used to evaluate contour integrals, especially when functions have singularities. It helps simplify real integrals, compute series, and is applied in many areas of science and engineering, including quantum physics.

Complex Functions and Singularities

Holomorphic functions, also known as analytic functions, are differentiable functions at every point in an open subset of the complex plane. A function f(z) is holomorphic in a region if it has a complex derivative f′(z) at every point in that region.

Singularities are points where a function ceases to be holomorphic. They are classified into three types:

**Removable Singularities: At a removable singularity, a function f(z) can be redefined at the point to make it holomorphic. For example, sinz/z has a removable singularity at z = 0 because sinz/z approaches 1 as z approaches 0.

**Poles: A pole is a point where a function goes to infinity. Poles can be:

• **Simple Poles: If \lim_{z \to c} (z-c)f(z) is finite and non-zero, then z = c is a simple pole.
• **Multiple Poles: If \lim_{z \to c} (z-c)^n f(z) is finite and non-zero for some integer n > 1, then z = c is a pole of order n.

**Essential Singularities: At an essential singularity, the behaviour of the function is chaotic. An example is f(z)~=~e^{1/z}~at~z~=~0

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis for evaluating contour integrals. The Residue Theorem states that if a function f(z) is analytic inside and on a simple closed contour C, except for a finite number of isolated singularities inside C, then the integral of f(z) around C is 2πi times the sum of the residues of f at those singularities.

Mathematically, this is explained as:

\oint_C f(z) \, dz~=~2\pi i \sum \text{Res}(f, z_k)

where,

• Res(f,zk) denotes Residue of f at Singularity zk

Conditions for the theorem are:

• **Function Analyticity: f(z) must be analytic on and inside the contour C, except at isolated singularities.
• **Isolated Singularities: The singularities within C must be isolated.
• **Closed Contour: C must be a closed contour.

Proof of Residue Theorem

The proof of the Residue Theorem involves deforming the contour and applying Cauchy's theorem. By considering small circles around the singularities and larger contours avoiding them, we can sum the residues at the singularities.

**Steps of Proof of Residue Theorem

**Step 1: Consider a function f(z) with isolated singularities z1, z2,…, zn inside a contour C. Deform the contour to avoid these singularities.

**Step 2: According to Cauchy's theorem, the integral over a contour that encloses no singularities is zero.

**Step 3: The integral around C can be broken into smaller integrals around each singularity: \oint_C f(z) \, dz = \sum_{k=1}^n \oint_{C_k} f(z) \, dz

**Step 4: \oint_{C_k} f(z) \, dz = 2\pi i \text{Res}(f, z_k)

**Step 5: Summing these integrals gives the final result: \oint_C f(z) \, dz = 2\pi i \sum_{k=1}^n \text{Res}(f, z_k)

Calculation of Residues

**Residues at Simple Poles

For a simple pole at z = c:

Res(f, c) = \lim_{z \to c} (z - c) f(z)

**Residues at Higher-Order Poles

For a pole of order n at z = c, the residue is obtained using the Laurent series expansion:

Res(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}} \left[(z-c)^n f(z)\right]

Applications of Residue Theorem

The application of the residue theorem is as follows:

• **Evaluating Real Integrals: The Residue Theorem of Calculus is quite useful for the evaluation of real integral notations; especially with trigonometric or exponential functions.
• **Solving Differential Equations: Residue calculus proves essential in finding particular solutions of differential equations which have complex coefficients.
• **Laplace Transforms: It aids in evaluating Laplace transforms by providing residues at the poles of the transformed function.
• **Electromagnetic Theory: In electromagnetics, the theorem is used to evaluate integrals representing physical quantities.
• **Fluid Dynamics: Residue calculus helps in solving potential flow problems in fluid dynamics.

Examples on Residue Theorem

**Example 1: Calculate the residue of f(z) = ez/(z2 + 1) at z = i.

**Solution:

f(z) = \frac{e^z}{(z-i)(z+i)}

At z = i, the residue is:

Res(f, i) = \lim_{z \to i} (z - i) \frac{e^z}{(z-i)(z+i)} = \frac{e^i}{2i}

**Example 2: Calculate the residue of f(z) = 1/(z – 1)3 at z = 1.

**Solution:

f(z) = \frac{1}{(z-1)^3}

Since it is a pole of order 3, the residue is:

Res(f, 1) = \frac{1}{2!} \lim_{z \to 1} \frac{d^2}{dz^2} \left[(z-1)^3 \frac{1}{(z-1)^3}\right] = \frac{1}{2} \cdot 2! = 1

**Example 3: Evaluate the integral: \oint_{C} ez/(z2 + 1) dz where, C is circle |z| = 2.

Solution:

ez/(z2 + 1) has singularities where z2 + 1 = 0

z = i and z = -i

Both singularities i and -i lie within the contour ∣z∣ = 2

Residue at z = i

Res{ ez/(z2 + 1), i} = limz→i(z - i){ez/(z - i)(z + i)}

= limz→i{ez/(z + i)}

= ei/2i

Residue at z = -i

Res{ ez/(z2 + 1), -i} = limz→-i(z + i){ez/(z - i)(z + i)}

= limz→-i{ez/(z - i)}

= e-i/(-2i)

= -e-i/(2i)

Apply the Residue Theorem

\oint_{C}ez/(z2 + 1) dz = 2πi{ei/2i + (-e-i/2i)}

= π{ei - e-i}

= π(2isin1) = 2πisin1

Unsolved Questions on Residue Theorem

**Question 1: Compute \int_{-\infty}^{\infty}\frac{dx}{x^2+1} using residues.

**Question 2: Evaluate \int_{-\infty}^{\infty}\frac{x^2}{(x^2+1)(x^2+4)}dx.

**Question 3: Calculate the residue of f(z) = sin z / z3 .

**Question 4: Use residues to evaluate \int_{0}^{\infty}\frac{x^p}{1+x^2}\,dx for −1 < p < 1.