Cauchy Theorem (original) (raw)

Last Updated : 12 Nov, 2025

**Cauchy's Theorem states that if a function is analytic within a closed contour and its interior, then the integral of that function around the contour is zero. This fundamental result is central to complex analysis, providing insights into the behavior of analytic functions in complex domains.

Cauchy's Theorem Statement

Cauchy's Integral Theorem states that

If f(z) is analytic throughout a simply connected region containing a closed contour C, then the integral of f(z) around C is equal to zero.

Simply Connected Region

A simply connected region is a region where any loop can be continuously contracted to a point without leaving the region, and any two points can be connected by a path within the region.

A region D in the complex plane (or in any topological space) is **simply connected if it has two properties:

• **Path-Connected: Any two points in D can be connected by a continuous path that lies entirely within D.
• **No Holes: Any closed loop (a continuous function from a circle into the region) in D can be continuously shrunk to a single point within D. This means there are no "holes" in the region.

Mathematical Formulation for Cauchy's Theorem

The mathematical formulation of Cauchy's Integral Theorem is expressed as follows:

**∮(f(z) dz) = 0

Where:

• ∮ denotes the contour integral around a closed contour.
• f(z) is an analytic function defined within and on the contour.
• dz represents an infinitesimal displacement along the contour.

**Related Articles:

• Calculus in Maths
• Derivatives
• Differential Calculus

Cauchy’s Integral Formula

Cauchy's Integral Formula states that if f(z) is analytic inside a simple closed contour (C), and (z0) is any point inside (C), then for any (n)th derivative of f(z), the formula is:

f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} \, dz

Where,

• f(n)(z0​) represents the nth derivative of f(z) evaluated at z0​
• n! denotes the factorial of n
• ∮C​ signifies the contour integral along the closed contour C
• z is a complex variable
• z0​ is a point within the contour C

Generalization of Cauchy’s Integral Formula

The generalized Cauchy Integral Formula extends the original formula—which gives only the first derivative—to compute higher-order derivatives of an analytic function. It allows us to find the 2nd, 3rd, … nth derivative of a function at any point inside a contour using a contour integral. This is very useful in complex analysis and in applications across physics and engineering.

f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} \, dz

Applications of Cauchy Theorem

Cauchy's Integral Theorem has numerous applications across various fields of mathematics and physics. Some of the key applications include

• **Complex Analysis: It helps understand complex functions and compute contour integrals.
• **Residue Theorem: It enables computation of real integrals using complex analysis.
• **Harmonic Functions: Studies solutions to Laplace's equation, used in electrostatics and fluid dynamics.
• **Conformal Mapping: It transforms complex domains while preserving angles, used in cartography and image processing.
• **Potential Theory: It solves problems related to gravitational and electrostatic potentials.
• **Engineering and Physics: It is applied in electromagnetism, fluid mechanics, and quantum mechanics for problem-solving.

Cauchy's Residue Theorem

Cauchy's Residue Theorem is a fundamental result in complex analysis that provides a powerful method for computing contour integrals of functions with singularities. It states that if f(z) is analytic inside and on a simple closed contour C, except at a finite number of isolated singular points z1​,z2​,...,zn​, then the contour integral of f(z) around C is equal to 2πi times the sum of the residues of f(z) at the singular points inside C. Mathematically, it can be expressed as:

\oint_C f(z) \, dz = 2\pi i \sum_{k=1}^n \text{Res}(f, z_k)

Where:

• ∮C​ denotes the contour integral around the closed contour C
• f(z) is a function analytic inside and on C, except at isolated singular points z1​,z2​,...,zn​
• (f, zk​) represents the residue of f(z) at the singular point zk​

Extension of Cauchy's Theorem

Cauchy-Goursat Theorem is an extension of Cauchy's Integral Theorem for simply connected regions in complex analysis. It states that if a function f(z) is analytic within a simply connected region D and its contour, then the integral of f(z) around any closed contour C within D is zero.

Mathematically, the Cauchy-Goursat Theorem can be stated as follows:

**∮ C **​f (z) dz = 0

Where:

• ∮C​ denotes the contour integral around the closed contour C
• f(z) is an analytic function within the simply connected region D and its contour

Solved Examples on Cauchy Theorem

**Example 1: Given a function f(z) that is analytic within a simply connected region D and its contour C, compute the contour integral ∮C​f(z)dz.

**Solution:

To solve the contour integral ∮C​f(z)dz, where f(z) is analytic within a simply connected region D and its contour C, we can apply Cauchy's Integral Theorem.

According to Cauchy's Integral Theorem, if f(z) is analytic within D and its contour C, then the contour integral ∮C​f(z).dz is equal to zero.

Therefore, solution to given problem is:

**∮ C ​f(z)dz=0

Result holds true for any simply connected region D and its contour C where f(z) is analytic.

**Example 2: A function f(z) has a simple pole at z=2 and a removable singularity at z=−1. Calculate the residue of f(z) at each singularity.

**Solution:

To calculate the residue of f(z) at each singularity, we need to determine the coefficient of the term \frac{1}{z - z_0} ​ in the Laurent series expansion of f(z) around each singularity z0​.

**Simple Pole at z = 2

Since f(z) has a simple pole at z = 2, the residue Res(f,2) is given by the coefficient of \frac{1}{z - 2} in the Laurent series expansion of f(z) around z = 2.

Let's denote g(z) = \frac{1}{z - 2}​. Then, the Laurent series expansion of g(z) around z=2 is simply g(z) = \frac{1}{z - 2}​.Since f(z) has a simple pole at z = 2, the residue Res(f, 2) is equal to the coefficient of \frac{1}{z - 2} in the Laurent series expansion of f(z) around z = 2.

Therefore, Res(f, 2) = coefficient of \frac{1}{z - 2}​ in f(z)

**Removable Singularity at z = −1

Since f(z) has a removable singularity at z=−1, the residue Res(f,−1) is zero because there is no pole at z=−1. In general, for a function with a removable singularity, the residue at that singularity is zero.

So, to summarize:

• Residue of f(z) at the simple pole z=2 is the coefficient of \frac{1}{z - 2}​ in f(z)'s Laurent series expansion around z = 2
• Residue of f(z) at the removable singularity z = −1 is zero

**Example 3: Using Cauchy's Integral Formula, find the value of the function \ f(z) = \frac{1}{z^2(z-1)} at z = 0

**Solution:

To find the value of the function f(z) = \frac{1}{z^2(z-1)}​ at z = 0 using Cauchy's Integral Formula, we first need to ensure that the function is analytic within a simply connected region containing the point z=0 and its contour.

Given that function has poles at z = 0 and z = 1, and z = 0 is the point of interest, we'll exclude z = 1 and its neighborhood from our consideration to ensure a simply connected region.

Now, let's apply Cauchy's Integral Formula for derivatives

f(z) = \frac{1}{z^2(z-1)}

Function has a simple pole at z = 0, so we'll use the first-order derivative of f(z) in the formula

f'(z) = \frac{d}{dz}\left(\frac{1}{z^2(z-1)}\right)

= \frac{d}{dz}\left(\frac{1}{z^2}\right) \cdot \frac{1}{z-1} + \frac{1}{z^2} \cdot \frac{d}{dz}\left(\frac{1}{z-1}\right)

= -\frac{2}{z^3} \cdot \frac{1}{z-1} - \frac{1}{z^2} \cdot \frac{d}{dz}\left(\frac{1}{z-1}\right)

= -\frac{2}{z^3} \cdot \frac{1}{z-1} - \frac{1}{z^2} \cdot \frac{-1}{(z-1)^2}

Now, we'll use Cauchy's Integral Formula

f'(0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{(z - 0)^2} \, dz

= \frac{1}{2\pi i} \oint_C \frac{\frac{1}{z^2(z-1)}}{z^2} \, dz

= \frac{1}{2\pi i} \oint_C \frac{1}{z^4(z-1)} \, dz

= \frac{1}{2\pi i} \times 2\pi i \times \text{Res}(f, 0)

= Res(f, 0)

So, f′(0) = Res(f, 0)

Now, to find the residue at z = 0, we need to find the coefficient of 1/z​ in the Laurent series expansion of f(z) around z=0.

f(z) = \frac{1}{z^2(z-1)} = \frac{A}{z} + \frac{B}{z^2} + \frac{C}{z-1}

Multiplying both sides by z2(z−1), we get:

1 = A(z−1) + Bz + Cz2

At z = 0, 1 = −A, so A=−1

Therefore, Res(f,0) = −1

Now, let's find f ′(0)

f ′(0) = −1

So, the value of f(z) at z = 0 is f(0) =f′(0) =−1.

Practice Problems on Cauchy Theorem

**Question 1: Prove Cauchy's Integral Theorem for a function f(z) that is analytic within a closed contour ( C ) and its interior.

**Question 2: Evaluate the contour integral (\oint_C \frac{\sin z}{z} \, dz ) where ( C ) is the unit circle centered at the origin.

**Question 3: Apply Cauchy's Residue Theorem to compute the integral ( \oint_C \frac{e^z}{z^2 + 1} \, dz ) where ( C ) is the contour ( |z| = 2 ).

**Question 4: Find the residue of ( f(z) = \frac{e^z}{z^2 - 4} ) at each singularity, and use it to compute the integral ( ∮C​f(z), dz ) along the contour ( C ) where ( C ) is the circle ( |z| = 3 ).

**Question 5: Using Cauchy's Integral Formula, determine the value of \oint_C \frac{e^z}{z} \, dzwhere ( C ) is the contour given by the line segment from z = 1 to z = 2i, followed by the line segment from z = 2i to z = -1.