Problem 25 - Advanced Engineering Mathematics

Question

Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. \(\oint_{C} \frac{z e^{z}}{z^{2}-1} d z, C:|z|=2\)

Step-by-Step Solution

Answer

The integral evaluates to \( 2\pi i (e - e^{-1}) \).

Step 1: Identify Singularities

The given function is \( f(z) = \frac{z e^z}{z^2 - 1} \). We find that the singularities are where the denominator is zero, i.e., \( z^2 - 1 = 0 \). This gives us singularities at \( z = 1 \) and \( z = -1 \).

Step 2: Check Singularities Inside Contour

The contour \( C \) is the circle \(|z| = 2\). Both singularities, \( z=1 \) and \( z=-1 \), lie within this contour as their moduli are less than 2.

Step 3: Calculate Residue at \( z = 1 \)

The function \( f(z) \) can be written as \( f(z) = \frac{e^z}{z+1} \) near \( z = 1 \), where the denominator \( z-1 \) gives a simple pole. The residue at \( z = 1 \) is given by \( \lim_{z \to 1} (z-1)f(z) = e \).

Step 4: Calculate Residue at \( z = -1 \)

Similarly, rewriting \( f(z) = \frac{e^z}{z-1} \) near \( z = -1 \), we find a simple pole at \( z = -1 \). The residue here is \( \lim_{z \to -1} (z+1)f(z) = -e^{-1} \).

Step 5: Apply Cauchy's Residue Theorem

According to Cauchy's residue theorem, the integral \( \oint_{C} f(z) \ dz \) is \( 2\pi i \) times the sum of the residues within \( C \). Thus, \( 2\pi i (e - e^{-1}) \) is the value of the integral.

Key Concepts

Singularities in Complex Analysis

In complex analysis, singularities refer to the points where a complex function is not analytic. This means the function cannot be expressed as a well-defined power series around these points.
A common type of singularity is the pole. If a function behaves like \( \frac{1}{(z-a)^n} \) near a point \( a \), \( a \) is called a pole of order \( n \).
Finding singularities involves identifying where the denominator of the function becomes zero, leading to undefined values. For instance, in the exercise, the function \( f(z) = \frac{z e^z}{z^2 - 1} \) has singularities at \( z^2 - 1 = 0 \), giving us \( z = 1 \) and \( z = -1 \).

• Simple poles are singularities where the function looks like \( \frac{1}{z-a} \) near the point \( a \).
• Calculating residues involves finding the coefficient of \( \frac{1}{z-a} \) in the Laurent series expansion of the function near the singularity.

Recognizing and understanding singularities allows us to effectively apply tools like Cauchy's Residue Theorem.

Contour Integration

Contour integration is a method used in complex analysis to evaluate integrals of complex functions over certain paths or curves in the complex plane. These curves are known as contours.
The main advantage of contour integration is that it allows us to compute integrals around singularities using contour paths without needing to evaluate indefinable expressions directly. In the exercise, the contour is \( |z| = 2 \), a circle of radius 2 centered at the origin.
This circular contour encloses the singularities \( z = 1 \) and \( z = -1 \), which are both within this boundary.

• Selecting the correct contour is crucial, as it should ideally encompass the singularities whose residues contribute to the integral's value.
• The process involves checking which singularities lie inside the chosen contour, ensuring that the integration path accounts for them accordingly.

Through contour integration, seemingly difficult real integrals become manageable by converting them into complex integrals.

Complex Analysis and Cauchy's Residue Theorem

Complex analysis is the study of complex numbers and functions of a complex variable, offering techniques for solving integrals, differential equations, and more. One of its key tools is Cauchy's Residue Theorem.
The theorem states that the integral of a function around a closed contour is \( 2\pi i \) times the sum of the function's residues within that contour.
In the exercise, after identifying the singularities and calculating the residues at \( z = 1 \) and \( z = -1 \), the residue theorem asserts that the integral is \( 2\pi i (e - e^{-1}) \).

• The residues are calculated for each singularity individually, as demonstrated in the solution steps.
• Finally, these values are summed up to find the integral along the contour.

Understanding complex analysis and particularly Cauchy's Residue Theorem simplifies evaluating integrals, especially in cases involving complex functions with singularities.