Problem 18 - Advanced Engineering Mathematics

Question

In Problems 17-20, use Cauchy's residue theorem, where appropriate, to evaluate the given integral along the indicated contours. \(\oint_{C} \frac{z+1}{z^{2}(z-2 i)} d z\) (a) \(|z|=1\) (b) \(|z-2 i|=1\) (c) \(|z-2 i|=4\)

Step-by-Step Solution

Answer

(a) \(\pi\), (b) \(\frac{\pi}{2} + \frac{\pi i}{2}\), (c) \(\frac{\pi}{2} + \pi^2\)."

Step 1: Identifying Singularities

The integrand function is \( \frac{z+1}{z^{2}(z-2i)} \). The singularities are at \( z=0 \) (which is a pole of order 2) and \( z=2i \) (which is a simple pole).

Step 2: Analyze Contour (a) |z| = 1

The contour \(|z|=1\) encloses the circle of radius 1 centered at the origin. This contour includes the pole at \( z=0 \) but not \( z=2i \), since \(|2i|=2\).

Step 3: Residue Calculation at z = 0

For the pole at \( z=0 \), which is an order 2 pole, the residue is calculated using:\[ \text{Residue at } z=0 = \lim_{z \to 0} \frac{d}{dz} \left( z^2 \cdot \frac{z+1}{z^2(z-2i)} \right) = \lim_{z \to 0} \frac{z+1}{z-2i} = \frac{1}{-2i} = \frac{-i}{2}. \]

Step 4: Evaluating Integral for |z| = 1

Using the residue theorem, the integral is \( 2 \pi i \times \text{Residue at } z=0 \):\[ \oint_{|z|=1} \frac{z+1}{z^{2}(z-2i)} d z = 2 \pi i \times \frac{-i}{2} = \pi. \]

Step 5: Analyze Contour (b) |z-2i| = 1

The contour \(|z-2i|=1\) encloses a circle of radius 1 centered at \(2i\). This contour includes the pole at \( z=2i \) but not \( z=0 \), since the contribution from \( z=0 \) is outside the circle.

Step 6: Residue Calculation at z = 2i

The residue at the simple pole \( z=2i \) is given by:\[ \text{Residue at } z=2i = \lim_{z \to 2i} (z-2i) \cdot \frac{z+1}{z^{2}(z-2i)} = \frac{(2i+1)}{(2i)^2} = -\frac{1}{4} + \frac{i}{4}. \]

Step 7: Evaluating Integral for |z-2i| = 1

Using the residue theorem, the integral is \( 2 \pi i \times \text{Residue at } z=2i \):\[ \oint_{|z-2i|=1} \frac{z+1}{z^{2}(z-2i)} d z = 2 \pi i \times \left( -\frac{1}{4} + \frac{i}{4} \right) = \frac{\pi}{2} + \frac{\pi i}{2}. \]

Step 8: Analyze Contour (c) |z-2i| = 4

The contour \(|z-2i|=4\) encloses a circle of radius 4 centered at \(2i\). This contour includes both poles at \( z=0 \) and \( z=2i \).

Step 9: Total Residue for |z-2i| = 4

The total residue for the contour is the sum of residues at \( z=0 \) and \( z=2i \):\[ \frac{-i}{2} + \left( -\frac{1}{4} + \frac{i}{4} \right) = -\frac{1}{4} - \frac{i}{4} - \frac{i}{4} = -\frac{1}{4} - \frac{i}{2}. \]

Step 10: Evaluating Integral for |z-2i| = 4

Using the residue theorem, the integral is \( 2 \pi i \times \text{Total Residue} \).\[ \oint_{|z-2i|=4} \frac{z+1}{z^{2}(z-2i)} d z = 2 \pi i \times \left(-\frac{1}{4} - \frac{i}{2}\right) = \frac{-\pi}{2} \times i - 2 \pi^2 \times i. \] The final calculation simplifies to: \[ \frac{\pi}{2} + \pi^2 \].

Key Concepts

Complex Analysis

Complex analysis is a fascinating area of mathematics that deals with complex numbers and functions of a complex variable. A complex number is a number of the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, defined by \(i^2 = -1\). In complex analysis, we study the properties and behaviors of functions that map complex numbers to complex numbers.
Some key concepts in complex analysis include:

• Complex functions: These are functions where the input and output are both complex numbers. An important class of such functions is analytic functions, which are differentiable in terms of complex variables and have many beautiful properties.
• Analyticity: This indicates that a function has a derivative at every point in its domain, similar to the concept of differentiability in real analysis.

Complex analysis is not only crucial in its own right but also incredibly useful in physics, engineering, and applied mathematics. It provides powerful techniques for solving problems involving waves, quantum mechanics, and fluid dynamics.

Contour Integration

Contour integration is a powerful technique in complex analysis used to evaluate integrals of complex functions. It involves integrating a complex-valued function along a specific path, or contour, in the complex plane. This method is especially useful when traditional real-variable techniques do not apply or are too complicated.
Here's how contour integration generally works:

• Contours: These are paths in the complex plane, often specified as the boundary of a region, such as a circle or rectangle.
• Integral Evaluation: The contour integral of a function \(f(z)\) over a contour \(C\) is expressed as \( \oint_C f(z) dz \). Cauchy's theorem and the residue theorem are crucial results that simplify these evaluations under certain conditions.

In contour integration, it is essential to identify which singularities (if any) are enclosed by the contour, as they heavily influence the value of the integral. The residues at these singularities play a central role in calculating the integral's value through Cauchy's residue theorem.

Singularities in Complex Functions

Singularities are points at which a complex function ceases to be analytic, meaning the function does not have a derivative at that point. Understanding singularities is crucial in complex analysis because they often contribute heavily to the behavior of the function over a region.
There are different types of singularities:

• Poles: Poles are isolated singularities where a function grows infinitely large. For example, in the function \( \frac{1}{z} \), the point \( z = 0 \) is a simple pole of order 1, because the function behaves like \( \frac{1}{z} \).
• Essential singularities: Essential singularities are more complicated points of non-analyticity where the function exhibits erratic behavior, without being predictable.
• Removable singularities: These occur where the function can be "redefined" or "completed" to be analytic. An example is the function \( \frac{\sin(z)}{z} \), which can be defined at \( z = 0 \) to make it analytic everywhere.

Determining the nature of singularities helps in applying the residue theorem for contour integration, as the residues are calculated based on the type and location of the poles within the chosen contour. This plays a significant role in solving complex integrals efficiently.