Problem 20 - Advanced Engineering Mathematics

Question

Use Cauchy's residue theorem, where appropriate, to evaluate the given integral along the indicated contours. \(\oint_{C} \frac{1}{z \sin z} d z\) (a) \(|z-2 i|=1\) (b) \(|z-2 i|=3\) (c) \(|z|=5\)

Step-by-Step Solution

Answer

(a) 0, (b) 0, and (c) sums of residues of poles within \(|z|=5\)

Step 1: Understanding the Integral and Contour

We need to evaluate the contour integral \( \oint_{C} \frac{1}{z \sin z} \, dz \) along different contours specified by \(|z-2i|=1\), \(|z-2i|=3\), and \(|z|=5\). To solve this, we will apply Cauchy's Residue Theorem, which involves finding the residues of poles inside the given contour. Since \(\sin z\) has zeros at \(z = n\pi\) (\(n\) is an integer), these zeros become poles in the integrand.

Step 2: Identifying the Poles

The function \( \frac{1}{z \sin z} \) has simple poles at \( z = 0 \) (since \(z=0\) makes the denominator zero), and at \( z = n \pi \) for \( n eq 0 \) because \( \sin(n\pi) = 0 \). Additionally, these poles repeat for each \(n\) along the real axis at integer multiples of \(\pi\). We need to consider which of these poles lie within the contours \(|z-2i|=1\), \(|z-2i|=3\), and \(|z|=5\).

Step 3(a): Evaluating the Integral for |z-2i|=1

The contour \(|z-2i|=1\) is a circle centered at \(2i\) with radius 1. To see if there are any poles within this circle, note that only poles that are closer than 1 unit to \(2i\) are inside. Here, no poles \( z = 0, n\pi \) (\(n\) being any integer makes \(z=n\pi\) entirely along the real line and hence outside \(|z-2i|=1\)), therefore, the integral evaluates to zero by Cauchy’s theorem as there are no poles inside the contour.

Step 3(b): Evaluating the Integral for |z-2i|=3

The contour \(|z-2i|=3\) is a circle centered at \(2i\) with radius 3. For this contour, poles on the imaginary line like \(z=0\) or near such as \(z=\pi\) or \(z=-\pi\) are still not within 3 units vertical of \(2i\). Thus, no poles of \(z=n\pi\) fall within \(|z-2i|=3\), hence the integral is also zero as there are no poles present within the contour.

Step 3(c): Evaluating the Integral for |z|=5

The contour \(|z|=5\) is a circle of radius 5 centered at the origin. For this contour, the pole at \(z = 0\) is within this circle, as are poles at small integer multiples of \(\pi\) such as \(z = \pm\pi, \pm2\pi, ...\) depending on the modulus. In particular, the relevant poles inside this circle up to \(z=5\) are \(z=0, \pm\pi\). The residues at these poles can be calculated individually and then summed up to evaluate the integral using the residue theorem.

Step 4: Calculating the Residues

For the pole at \(z=0\), the residue can be calculated as the limit: \( \lim_{z \to 0} z \cdot \frac{1}{z \sin z} = \frac{1}{0 \cdot \sin 0} \), which simplifies using L'Hospital's rule, gives \(\frac{1}{1} = 1\). Similarly, for poles like \(z = \pi\), the residue is \( \lim_{z \to \pi} (z - \pi) \cdot \frac{1}{z \sin z} = \frac{1}{\pi \cdot 0} \) gives \(-\frac{1}{\pi}\). Sum these residues for the total but ensuring the counter interpretation.

Step 5: Applying the Residue Theorem

By the residue theorem, the integral around the contour \(|z|=5\) is \(2\pi i\) times the sum of the residues inside the contour. Sum up the calculated values for poles \(z=0, \pm\pi, \pm2\pi\) etc., and multiply by \(2\pi i\) to obtain the result of the integral.

Key Concepts

Contour Integration

Contour integration is a method used in complex analysis to evaluate complex integrals over a path or contour in the complex plane. It's a powerful technique since it considers both the path (or contour) through which the integration runs and the singularities, or points of interest, inside the contour.

Some key aspects of contour integration include:

• Contours: These are paths defined in the complex plane. They are often circles or paths where some property or distance is constant, such as \(|z-2i|=3\), which describes a circle centered at point \(2i\) with a radius of \(3\).
• Direction: The direction in which the contour is traversed matters. Most of the time, integration is performed counterclockwise unless otherwise specified.
• Application of Theorems: Theorems, like Cauchy's integral theorem, simplify complex integral calculations by considering the contour and the nature of singularities inside it.

Understanding contour integration is essential when working within complex analysis because it allows us to evaluate integrals that might seem unsolvable using standard real calculus methods.

Complex Analysis

Complex analysis is a branch of mathematics exploring functions of complex numbers. It builds upon calculus treating complex numbers and extends many ideas found in real analysis into the complex domain.

Here are some interesting insights about complex analysis:

• Complex Numbers: These come in the form \(a + bi\), where \(a\) is the real part and \(b\)i is the imaginary part.
• Holomorphic Functions: Functions that are analytic and differentiable everywhere in their domain. Most contour integrals involve these functions since they are continuous and smooth.
• Cauchy's Theorems: Fundamental to complex analysis, they simplify contours integration problems significantly, especially Cauchy's residue theorem, which relates contour integrals to residues of poles within the contour.

Complex analysis not only provides tools for evaluating integrals but also opens doors to deeper understanding of mathematical properties and relationships in the complex plane.

Residue Calculation

Residue calculation is a critical part of solving contour integrals using Cauchy's residue theorem. In the context of complex functions, residues are the coefficients of the term \((z - z_0)^{-1}\) in the Laurent series expansion of a function.

To understand residue calculation, consider these aspects:

• Identifying Poles: Essential to finding residues. Poles are values of \(z\) where the function becomes undefined, typically leading to infinite values.
• Simple Poles: These are poles of order \(1\), easily handled by residue calculations. For instance, in the function \(\frac{1}{z \, \sin z}\), poles exist where \(\sin z = 0\), namely at multiples of \(\pi\).
• Residue Formula: For a simple pole at \(z = z_0\), the residue \(\text{Res}(f, z_0)\) can be found using the formula: \(\lim_{z \to z_0} (z - z_0)f(z)\).

Once residues are calculated for all poles within a contour, the integral can be determined by summing these residues and multiplying by \(2\pi i\), thanks to the residue theorem.