Problem 6 - Complex Analysis for Mathematics and Engineering

Question

Use residues to find $$ \int_{0}^{2 \pi} \frac{\sin ^{2} \theta}{5-3 \cos \theta} d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{2\pi}{3}\).

Step 1: Convert to Complex Form

To use residues, we first convert the trigonometric integral to a contour integral in the complex plane. We use the substitutions \(z = e^{i\theta}\), \(d\theta = \frac{dz}{iz}\), \(\cos \theta = \frac{z + z^{-1}}{2}\), and \(\sin \theta = \frac{z - z^{-1}}{2i}\). Thus, \(\sin^2 \theta = \left(\frac{z - z^{-1}}{2i}\right)^2 = \frac{(z^2 - 2 + z^{-2})}{-4}\).

Step 2: Simplify and Setup Integral

Substitute \(\sin^2 \theta\) and \(\cos \theta\) into the integral:\[\int_0^{2\pi} \frac{\sin^2 \theta}{5 - 3 \cos \theta} \, d\theta = \oint_{|z|=1} \frac{- (z^2 - 2 + z^{-2})/4}{5 - \frac{3}{2}(z + z^{-1})} \cdot \frac{dz}{iz}\]After simplifying, we get:\[\oint_{|z|=1} \frac{z^4 - 2z^2 + 1}{z^3(6z - 3(z^2+1))} \, dz\]

Step 3: Find Residues

The poles of the integrand are at the zeros of the denominator inside the unit circle. The denominator is \(z^3(6z - 3(z+z^{-1})) = 3z^3(z-1)(z-\frac{1}{3})\). Only \(z = \frac{1}{3}\) is inside the unit circle. Calculate the residue at \(z = \frac{1}{3}\). This amounts to evaluating the residue at this simple pole, using: \[\text{Res}(f,z=\frac{1}{3}) = \lim_{z \to \frac{1}{3}} (z - \frac{1}{3})f(z)\]

Step 4: Evaluate Integral Using Residue Theorem

We apply the residue theorem which states that the contour integral of a function around a closed curve is \(2\pi i\) times the sum of residues inside the curve. In this case:\[\oint_{|z|=1} f(z) \, dz = 2\pi i \cdot \text{Res}(f, z = \frac{1}{3})\]Calculate the residue and multiply by \(2\pi i\) to evaluate the integral.

Key Concepts

Contour Integration

Contour integration is a method used in complex analysis to evaluate certain types of integrals, particularly those with a periodic or circular nature. It involves integrating a complex function along a specified contour in the complex plane. In this exercise, we transformed a real integral into a contour integral by representing the trigonometric function in terms of complex variables.
To fully utilize contour integration, one typically completes the following steps:

• Transform the real integral into a contour integral using appropriate substitutions. For trigonometric integrals, this often involves substituting trigonometric functions with their complex exponential forms.
• Identify a closed contour that simplifies the integral. A common choice is the unit circle, represented by \(|z| = 1\).
• Use the residue theorem to evaluate the contour integral, which involves determining the residues of singularities inside the contour.

By converting the real trigonometric integral to the contour integral \(\oint_{|z|=1} f(z) \, dz\), and with the residue theorem, what might initially appear complex can be significantly simplified. This approach is powerful in electrical engineering or physics applications, where periodic signals are prevalent.

Complex Analysis

Complex analysis is a branch of mathematics that studies functions of complex numbers. It has profound applications in many fields, including engineering, physics, and number theory. One powerful tool within complex analysis is the residue theorem, which provides an efficient way to evaluate contour integrals.
In the context of this exercise, complex analysis simplifies evaluating the integral of a trigonometric function over a periodic interval by converting it into a problem in the complex plane. Here's why this approach works:

• The use of complex numbers allows the representation of trigonometric identities in more manageable forms, via Euler's formula: \( z = e^{i\theta} \).
• This transformation condenses periodic integrals to contour integrals, making them easier to handle.
• Complex functions result in equations where singularities or poles within a contour can be identified, paving the way for using residues.

Ultimately, complex analysis with contour integration and the residue theorem turns challenging trigonometric integrals into elegant solutions, underscoring the real-world utility of mathematical abstraction.

Trigonometric Substitution

Trigonometric substitution is a technique for simplifying integrals involving trigonometric functions by substituting them with equivalent expressions. Here, trigonometric substitution helps bridge real integration and complex analysis.
For this problem, we used these key transformations:

• \( z = e^{i\theta} \): This substitution converts the variable from a real number to a complex exponential form.
• \( \cos \theta = \frac{z + z^{-1}}{2} \) and \( \sin \theta = \frac{z - z^{-1}}{2i} \): These relate trigonometric functions of \(\theta\) to expressions involving complex numbers \(z\).
• \( \sin^2 \theta = \frac{(z^2 - 2 + z^{-2})}{-4} \): This simplifies the function \(\sin^2 \theta\) into a polynomial form for easier integration.

Utilizing these substitutions, we transformed our integral, facilitating its conversion to a contour integral. In complex analysis, trigonometric substitution shines by reducing the complexity of periodic integrals, remaining a cornerstone understanding both in calculus and complex analysis spheres.