Problem 13 - Advanced Engineering Mathematics

Question

\(\oint_{C} \frac{z}{z^{2}-\pi^{2}} d z ;|z|=3\)

Step-by-Step Solution

Answer

The integral is \( 2\pi i \).

Step 1: Identify the poles inside the contour

The integrand, \( \frac{z}{z^{2}-\pi^{2}} \), has poles where the denominator is zero, i.e., where \( z^{2} - \pi^{2} = 0 \). This gives \( z = \pm \pi \). For the contour \(|z| = 3\), we identify which of these poles lie inside the contour. Since \(3 > \pi\), both poles \( z = \pi \) and \( z = -\pi \) are inside the contour.

Step 2: Apply the Residue Theorem

The Residue Theorem states that \( \oint_C f(z)\,dz = 2\pi i \times \) (sum of residues of \( f(z) \) inside \( C \)). We need to find the residues of \( f(z) = \frac{z}{z^2-\pi^2} \) at the poles \( z=\pi \) and \( z=-\pi \).

Step 3: Calculate the Residue at \( z = \pi \)

The residue of a function \( \frac{g(z)}{h(z)} \) at a simple pole \( z=z_0 \) is given by \( \frac{g(z_0)}{h'(z_0)} \). Here, \( g(z) = z \) and \( h(z) = z^2 - \pi^2 \). Differentiating \( h(z) \) gives \( h'(z) = 2z \). The residue at \( z = \pi \) is \( \frac{\pi}{2\pi} = \frac{1}{2} \).

Step 4: Calculate the Residue at \( z = -\pi \)

Similarly, the residue at \( z = -\pi \) is \( \frac{-\pi}{2(-\pi)} = \frac{1}{2} \).

Step 5: Sum the Residues and Compute the Integral

The sum of the residues inside the contour is \( \frac{1}{2} + \frac{1}{2} = 1 \). By the Residue Theorem, the contour integral is \( 2\pi i \times 1 = 2\pi i \).

Key Concepts

Contour Integration

Contour integration is a method used in complex analysis to evaluate integrals over a path or curve in the complex plane. This technique is a powerful tool when dealing with complex-valued functions. When performing contour integration:

• The path, known as the contour, can be any closed loop in the complex plane.
• The integral calculates a complex number that represents the sum of the products of the function values and infinitesimally small curve segments along the contour.

In the given problem, the contour is a circle defined by \( |z| = 3 \). The integral is performed over this circular path, around the origin. Contour integration is key in solving problems involving complex functions and is closely related to various theorems, such as Cauchy's Integral Theorem.

Residue Theorem

The Residue Theorem is a central concept in complex analysis, used for evaluating contour integrals of complex functions. It relies on the idea of residues, which are a measure of a function's behavior near its poles:

• A pole is a point where a function becomes unbounded or undefined.
• The residue is a specific coefficient that describes the behavior of the function close to a pole.

The theorem states that to evaluate \( \oint_C f(z) \, dz \), the integral of a function around a contour, you can use the formula \( 2\pi i \times \text{{(sum of residues inside the contour)}} \). In practical terms, calculating a contour integral involves determining the residues at the poles within the contour. In our example, the poles \( z = \pi \) and \( z = -\pi \) are both inside the contour \( |z| = 3 \,\) and their residues are summed to find the solution.

Complex Poles

Complex poles are specific points in the complex plane where a complex function becomes undefined due to division by zero in its rational form. Poles are critical in understanding the behavior of complex functions:

• A simple pole is a point where the function has a first-order zero in the denominator but is otherwise analytic or well-behaved.
• We calculate the residue of a simple pole using the formula \( \frac{g(z_0)}{h'(z_0)} \,\) where \(g(z)\) is the numerator and \(h(z)\) is the denominator of the function near the pole \(z_0\).

In our exercise, the poles occur when \( z^2 = \pi^2 \,\) yielding \( z = \pm \pi \.\) Both these poles are simple, meaning they can easily be handled with residue calculations, making the integral evaluation straightforward.

Integral Calculus

Integral calculus involves the computation of integrals and is a fundamental branch of mathematics, often applied to functions in the real and complex domains. In complex analysis, integral calculus has special tools and techniques due to the nature of complex functions:

• Complex integration handles functions of a complex variable along a specified contour.
• The evaluation often requires understanding the behavior of the function within and around the contour.

In the context of our problem, we use integral calculus to encompass the calculation of contour integrals and residues, ultimately solving for the total contour integral \( 2\pi i \) in the closed path \( |z| = 3 \,\) encapsulating our complex poles. Mastery of integral calculus in both real and complex scenarios allows mathematicians to solve a wide variety of theoretical and practical problems.