Problem 32 - Advanced Engineering Mathematics

Question

In Problems 21-32, use Cauchy's residue theorem to evaluate the given integral along the indicated contour. $$ \oint_{C} \frac{\cos z}{(z-1)^{2}\left(z^{2}+9\right)} d z, C:|z-1|=1 $$

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{2\pi i (3\sin 1 + 2\cos 1)}{10}\).\)

Step 1: Identify the poles

The function inside the integral is \( \frac{\cos z}{(z-1)^{2}(z^{2}+9)} \). The poles occur where the denominator is zero. So, first identify the poles: these are \( z = 1 \) (a pole of order 2) and \( z = \pm 3i \) (simple poles).

Step 2: Determine poles within the contour

The contour is defined by \(|z-1|=1\), which is a circle of radius 1 centered at \(z = 1\). The pole at \(z = 1\) lies within the contour, but the poles at \( z = 3i \) and \( z = -3i \) do not lie within this contour.

Step 3: Apply Cauchy's residue theorem

Cauchy's residue theorem states that \(\oint_C f(z) \, dz = 2\pi i \sum \text{Residues inside } C \). Since only \(z = 1\) is inside \(|z-1|=1\), we need to find the residue at this pole.

Step 4: Compute the residue at the pole z = 1 (order 2)

Since the pole at \(z = 1\) is of order 2, the residue is given by \(\lim_{z \to 1} \frac{d}{dz}\left((z-1)^{2} \cdot \frac{\cos z}{(z-1)^{2}(z^{2}+9)}\right) = \lim_{z \to 1} \frac{d}{dz}\left(\frac{\cos z}{z^{2}+9}\right)\).

Step 5: Differentiate and evaluate the limit

Calculate the derivative \( \frac{d}{dz}\left(\frac{\cos z}{z^{2}+9}\right) \) using the quotient rule: \( \left( (z^{2}+9)(-\sin z) - \cos z(2z) \right) / (z^{2}+9)^{2} \). Evaluate this limit as \( z \to 1 \), leading to the residue being \( \left(-3\sin 1 - 2\cos 1\right) / 10 \).

Step 6: Use the residue to evaluate the integral

The residue at \( z = 1 \) is \( \frac{-3\sin 1 - 2\cos 1}{10} \). According to Cauchy's residue theorem, the integral evaluates to \(2\pi i \) times this residue, which gives \(-\frac{2\pi i (3\sin 1 + 2\cos 1)}{10} \).

Key Concepts

Cauchy's Residue Theorem

Cauchy's Residue Theorem is a powerful tool in complex analysis that simplifies the evaluation of contour integrals. This theorem is particularly useful for functions with poles, which are points where the function becomes unbounded or undefined. The heart of Cauchy's Residue Theorem lies in its ability to evaluate integrals by considering only the residues, which are the coefficients of the \( \frac{1}{z} \) term in the Laurent series expansion of a function around its singularity (pole).

In simple terms, the theorem states:

• Given a contour \( C \) that encloses singularities (poles) of a function \( f(z) \), the integral around that contour can be calculated using the formula \( \oint_C f(z) \, dz = 2\pi i \sum \text{Residues inside } C \).
• This means that instead of directly computing complex integrals, we can sum up the residues of the poles enclosed by the contour and multiply by \( 2\pi i \).

To apply this theorem efficiently:

• Identify and locate the poles of the complex function.
• Determine which poles lie within the contour.
• Calculate the residue for each of these poles.

By focusing only on these residues, Cauchy's Residue Theorem significantly simplifies calculations that would otherwise be daunting.

Contour Integration

Contour integration refers to integrating a complex function along a given path or contour in the complex plane. It is a fundamental concept in complex analysis.

Contours are typically closed paths, such as circles or loops, that help define the boundaries around which the integral is taken. The beauty of contour integration is that it allows us to consider the influence of complex paths on integrals.

Key steps to carry out contour integration:

• Define the contour: A contour is often expressed as \( C: |z - a| = r \), where \( a \) is the center, and \( r \) is the radius of the circle.
• Ensure the function is analytic (smooth and differentiable) on and within the contour, except at the poles.
• Apply theorems like Cauchy's Integral Theorem or Residue Theorem, depending on the problem.

In our original problem, the contour was given as \( |z-1|=1 \, \) a circle centered at \( z = 1 \) with a radius of 1. This choice of contour plays a crucial role in simplifying the computation, as it helps us directly apply Cauchy's Residue Theorem.

Poles of Complex Functions

Poles are specific kinds of singularities in complex functions where the function becomes infinite. Understanding poles is crucial for evaluating integrals using techniques in complex analysis.

Here's what you need to know about poles:

• A pole of order \( n \) occurs at a point \( z_0 \), where around this point, the function behaves like \( \frac{1}{(z-z_0)^n} \).
• Simple poles are poles of order 1, where the function resembles \( \frac{1}{z-z_0} \).
• Higher order poles (like second order or third order poles) mean the function becomes more intense as it approaches that point.

To identify poles:

• Set the denominator of the given function to zero and solve for \( z \).
• Verify the order of each pole by examining how many times we need to factor out the singularity in the function.

In the exercise, the function \( \frac{\cos z}{(z-1)^{2}(z^{2}+9)} \) has a pole of order 2 at \( z = 1 \) and simple poles at \( z = \pm 3i \). Understanding where these poles are, and their nature is critical for determining which ones to include within the contour for the integration process.