Solved Examples on 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 [Tex]\frac{1}{z – z_0} [/Tex]​ 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 [Tex] \frac{1}{z – 2}[/Tex] in the Laurent series expansion of f(z) around z = 2.

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

Therefore, Res(f, 2) = coefficient of [Tex]\frac{1}{z – 2}[/Tex]​ 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 [Tex]\frac{1}{z – 2}[/Tex]​ 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 [Tex]\ f(z) = \frac{1}{z^2(z-1)}[/Tex] at z = 0

Solution:

To find the value of the function f(z) = [Tex]\frac{1}{z^2(z-1)}[/Tex]​ 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

[Tex]f(z) = \frac{1}{z^2(z-1)} [/Tex]

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

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

[Tex]= \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)[/Tex]

[Tex]= -\frac{2}{z^3} \cdot \frac{1}{z-1} – \frac{1}{z^2} \cdot \frac{d}{dz}\left(\frac{1}{z-1}\right)[/Tex]

[Tex]= -\frac{2}{z^3} \cdot \frac{1}{z-1} – \frac{1}{z^2} \cdot \frac{-1}{(z-1)^2}[/Tex]

Now, we’ll use Cauchy’s Integral Formula

[Tex]f'(0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{(z – 0)^2} \, dz[/Tex]

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

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

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

= 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.

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

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.

the Cauchy-GoursatCauchy’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.

In this article, we will learn about Cauchy’s Integral Theorem, Cauchy’s Integral Formula, It’s applications, Cauchy’s Residue Theorem and the Cauchy-Goursat Theorem.