Problem 6 - Advanced Engineering Mathematics

Question

Prove that \(\oint_{c} f(z) d z=0\), where \(f\) is the given function and \(C\) is the unit circle \(|z|=1\). \(f(z)=\frac{e^{z}}{2 z^{2}+11 z+15}\)

Step-by-Step Solution

Answer

The integral is zero because all poles of \(f(z)\) lie outside the unit circle.

Step 1: Identify Poles Inside the Contour

First, we need to find the poles of the function \(f(z) = \frac{e^z}{2z^2 + 11z + 15}\). The poles are the roots of the denominator, so we solve \(2z^2 + 11z + 15 = 0\) to find the poles.

Step 2: Solve the Quadratic Equation

Use the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots of \(2z^2 + 11z + 15\). Here, \(a = 2\), \(b = 11\), and \(c = 15\), so we solve it as: \[ z = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 2 \cdot 15}}{2 \cdot 2} \].

Step 3: Calculate the Discriminant

The discriminant \(\Delta = 11^2 - 4 \times 2 \times 15 = 121 - 120 = 1\). Thus, there are two distinct real roots for the quadratic equation.

Step 4: Find the Roots

Substitute the value of the discriminant into the quadratic formula: \[ z = \frac{-11 \pm 1}{4} \]. This gives us \(z_1 = \frac{-10}{4} = -2.5\) and \(z_2 = \frac{-12}{4} = -3\).

Step 5: Determine if Poles are Inside the Unit Circle

Both poles \(z_1 = -2.5\) and \(z_2 = -3\) have magnitudes greater than 1 (\(|z_1| = 2.5\) and \(|z_2| = 3\)), meaning they lie outside the unit circle \(|z| = 1\).

Step 6: Apply Cauchy's Integral Theorem

Since \(f(z)\) has no poles inside \(C\), Cauchy's Integral Theorem states that \(\oint_C f(z)\, dz = 0\) for a function that is analytic inside and on a simple closed contour.

Key Concepts

Contour Integration

Contour integration is a technique used in complex analysis for integrating complex functions along a specific path or contour in the complex plane. It differs from regular real-variable integration in that it traverses paths in the complex plane.

• A contour is a path in the complex plane composed of a finite number of smooth curves joined end to end.
• In contour integration, paths are usually closed loops, like the unit circle in the given problem, represented by \(|z|=1\).
• Calculating a contour integral involves evaluating the function along this path rather than just over an interval as in real analysis.

Contour integration provides powerful results, such as the residue theorem, and is essential in evaluating integrals that are otherwise difficult. It is a foundational tool in complex analysis.

Cauchy's Integral Theorem

Cauchy's Integral Theorem is a fundamental result in complex analysis that greatly simplifies the process of evaluating certain types of contour integrals.

• The theorem states that if a function is analytic everywhere inside and on some simple closed contour, like our unit circle \(|z| = 1\), then the integral of this function over that contour is zero, \(\oint_{C} f(z) \, dz = 0\).
• Analytic functions do not have singularities like poles within the integration path, which makes them integrable over a closed path without yielding non-zero results.
• This theorem is especially useful because it means we can ignore the path details as long as the function remains analytic within the region.

In our problem, we apply this theorem because all the poles, or singularities, of \(f(z)\) lie outside the unit circle, ensuring the integral is zero.

Analytic Functions

An analytic function, also known as a holomorphic function, is a complex function that is differentiable at every point in a region of the complex plane.

• Differentiability in the complex sense implies a stronger form of smoothness than in real analysis, meaning that the function not only has a derivative at each point but that the derivative itself behaves continuously.
• These functions can be expanded into power series (like Taylor series) within their radius of convergence, making them incredibly predictive within that region.
• In the context of contour integration and the Cauchy's Integral Theorem, being analytic guarantees the function can be integrated easily over closed paths without resulting in a non-zero value.

The given function \(f(z) = \frac{e^z}{2z^2 + 11z + 15}\) is analytic on and inside the contour as it lacks poles in that region, allowing us to apply the theorem confidently.

Poles of Complex Functions

Poles in complex analysis are a type of singularity where a complex function takes an infinite value. They are the points in the complex plane where the function becomes undefined, leading to potential issues in integrals.

• A pole of order \(n\) of a function \(f(z)\) is where \(f(z)\) behaves like \(\frac{1}{(z-z_0)^n}\) as \(z\) approaches \(z_0\).
• Poles are identified by looking at the roots of the function's denominator. In our given function, the poles are the solutions to \(2z^2 + 11z + 15 = 0\).
• If poles lie inside the contour of integration, they must be considered in the integral evaluation, often using residue theory. However, if outside, as in our problem, they don't affect the integral's value over the contour.

In this exercise, the poles \(z_1 = -2.5\) and \(z_2 = -3\) were found to be outside the unit circle, meaning the contour integral equals zero.