Problem 25 - A First Course in Complex Analysis with Applications

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 \( \pi i (e + e^{-1}) \).

Step 1: Identify Singularities

The singularities of the function \( f(z) = \frac{z e^{z}}{z^{2}-1} \) are the points where the denominator is zero. Solving \( z^{2} - 1 = 0 \) gives the singularities \( z = 1 \) and \( z = -1 \).

Step 2: Determine Singularities Inside the Contour

The contour \( C: |z|=2 \) is a circle of radius 2 centered at the origin. Both \( z = 1 \) and \( z = -1 \) lie inside this contour since their magnitudes are less than 2. Thus, these are the poles inside the contour.

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

The residue of \( \frac{z e^{z}}{z^{2}-1} \) at \( z=1 \) can be found using the formula for simple poles: \[ \text{Res}(f,1) = \lim_{z \to 1} (z-1)f(z) = \lim_{z \to 1} z e^{z} \times \frac{z-1}{(z-1)(z+1)} = \lim_{z \to 1} \frac{z e^{z}}{z+1} = \frac{1 \cdot e}{2} = \frac{e}{2}. \]

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

The residue of \( \frac{z e^{z}}{z^{2}-1} \) at \( z=-1 \) is calculated similarly:\[ \text{Res}(f,-1) = \lim_{z \to -1} (z+1) \frac{z e^{z}}{z^{2}-1} = \lim_{z \to -1} \frac{z e^{z}}{z-1} = \frac{-1 \cdot e^{-1}}{-2} = \frac{e^{-1}}{2}. \]

Step 5: Apply Cauchy's Residue Theorem

According to Cauchy's residue theorem, the integral around a closed contour is \( 2\pi i \) times the sum of the residues inside the contour. Therefore, \[ \oint_{C} \frac{z e^{z}}{z^{2}-1} dz = 2\pi i \left(\frac{e}{2} + \frac{e^{-1}}{2}\right) = \pi i (e + e^{-1}). \]

Key Concepts

Complex Analysis

Complex analysis is a fascinating field of mathematics that studies functions of complex numbers. These functions are distinguished by their dependency on a complex variable, usually denoted as \( z \), which can be expressed in terms of a real part and an imaginary part. The study does not just revolve around the functions themselves, but also considers their properties, manipulations, and implications in various domains.

One key feature of complex analysis is the study of

• analytic functions, which are infinitely differentiable
• singularities, which are points where a function doesn't behave well
• complex integration, which involves integrating functions along paths or contours in the complex plane

This study leads to powerful results and tools, such as Cauchy's Residue Theorem, which helps in evaluating complex integrals.

Contour Integration

Contour integration is a method of evaluating integrals of complex functions along a specified path or curve, known as a contour, in the complex plane. This method forms a major component of complex analysis and exploits the rich interplay between analytic functions and their contours.

In typical problems, the contour is often a simple geometric object like a circle or a line segment. Here, for instance, the contour was a circle \( C: |z| = 2 \), a common choice for enclosing singularities and simplifying the integration process. This problem focuses on integrating \( \frac{z e^{z}}{z^{2}-1} \) along this circular path.

The goal of contour integration often involves tackling integrals that would otherwise be extremely difficult to solve using real analysis techniques, relying instead on the properties of analytic functions and their descriptions in the complex plane.

Singularities

Singularities are special points where an otherwise well-behaved function becomes undefined or infinite. In complex functions, these occur at points where the function cannot be expressed as a power series. Understanding the location and nature of singularities is crucial for many complex analysis tasks.

There are different types of singularities:

• Poles: Points where the function heads towards infinity, like with \( \frac{1}{z} \) at \( z = 0 \)
• Essential Singularities: Points where the function behaves chaotically, such as \( e^{1/z} \) at \( z = 0 \)
• Branch Points: Points requiring multi-valued function definitions

In the exercise, the function \( \frac{z e^{z}}{z^{2}-1} \) has poles at \( z=1 \) and \( z=-1 \), found by setting the denominator equal to zero. These are found to be simple poles, as the function behaves well when multiplied by suitable factors.

Residue Calculation

Residue calculation is a core technique in evaluating complex integrals via Cauchy's Residue Theorem. Essentially, the residue is the coefficient of \( \frac{1}{z-a} \) in the Laurent series expansion of a function around a point \( a \). The residue offers a compact representation of a function's behavior near its singularities.

For simple poles, such as in this problem with poles at \( z = 1 \) and \( z = -1 \), calculating the residue involves multiplying the function by \( z - ext{(pole value)} \) and then simplifying. In our example:

• The residue at \( z = 1 \) is calculated as \( \frac{e}{2} \)
• The residue at \( z = -1 \) is calculated as \( \frac{e^{-1}}{2} \)

Cauchy's Residue Theorem then uses these residues to evaluate complex integrals: the integral of a function around a closed contour is equal to \( 2\pi i \) times the sum of the function's residues within the contour. This powerfully simplifies our task, allowing us to find the integral value without directly solving it.