Problem 26 - Advanced Engineering Mathematics

Question

In Problems 21-32, use Cauchy's residue theorem to evaluate the given integral along the indicated contour. $$ \oint_{C} \frac{e^{z}}{z^{3}+2 z^{2}} d z, C:|z|=3 $$

Step-by-Step Solution

Answer

The integral is \( \frac{\pi i}{2} (1 + e^{-2}) \).

Step 1: Simplify the integrand

First, factor the denominator of the integrand. We have: \[ z^3 + 2z^2 = z^2(z + 2). \] Hence, the function becomes \( \frac{e^z}{z^2(z+2)}. \)

Step 2: Identify singularities and residues

The function has singularities at \( z = 0 \) and \( z = -2 \). Both are within the contour \( C: |z| = 3 \). The singularity at \( z = 0 \) is of order 2, and the singularity at \( z = -2 \) is of order 1.

Step 3: Find the residue at $z=0$

For \( z = 0 \) (a pole of order 2), the residue is given by\[ \text{Res}(f, 0) = \lim_{z \to 0} \frac{d}{dz} \left( z^2 \cdot \frac{e^z}{z^2(z+2)} \right) = \lim_{z \to 0} \frac{d}{dz} \left( \frac{e^z}{z+2} \right). \]Differentiating yields:\[ \frac{d}{dz} \left( \frac{e^z}{z+2} \right) = \frac{e^z}{z+2} - \frac{e^z}{(z+2)^2}, \]evaluating at \( z=0 \):\[ \text{Res}(f, 0) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}. \]

Step 4: Find the residue at $z=-2$

For \( z = -2 \) (a simple pole), the residue is given as follows:\[ \text{Res}(f, -2) = \lim_{z \to -2} (z + 2) \cdot \frac{e^z}{z^2(z+2)} = \lim_{z \to -2} \frac{e^z}{z^2}. \]Evaluating this gives:\[ \text{Res}(f, -2) = \frac{e^{-2}}{4}. \]

Step 5: Apply Cauchy's Residue Theorem

Cauchy's Residue Theorem states that the integral of a function over a closed contour is \( 2\pi i \) times the sum of the residues within the contour. Thus,\[ \oint_{C} \frac{e^{z}}{z^{3}+2z^{2}} \, dz = 2\pi i \left( \frac{1}{4} + \frac{e^{-2}}{4} \right). \]Simplifying, we find:\[ \oint_{C} \frac{e^{z}}{z^{3}+2z^{2}} \, dz = \frac{\pi i}{2} (1 + e^{-2}). \]

Key Concepts

Complex Analysis

Complex analysis is a branch of mathematics that extends the principles of calculus to functions of complex numbers. Instead of working with real number systems, complex analysis delves into the realm of numbers composed of a real and an imaginary part.

A complex number is expressed as \( z = x + yi \), where \( x \) and \( y \) are real numbers and \( i \) is the imaginary unit, defined by \( i^2 = -1 \). Functions in complex analysis include these complex numbers.

A major benefit of complex analysis is that complex functions are infinitely differentiable where they are defined, which isn't always true for real functions. This advantage allows for powerful techniques such as contour integration and residue calculus.

Contour Integration

Contour integration is a method used in complex analysis for evaluating certain types of definite integrals, over one-dimensional curves in the complex plane. These curves are known as contours.

Imagine tracing a loop or path on the plane. This path is your contour. The integral we need to compute is carried out over this path. What makes contour integrals so powerful, especially with complex-valued functions, is that they account for a function's behavior along the entire path.

• You move from a starting point and trace out a loop or line, returning to the start.
• Complex paths can circle around singularities (points where a function isn't well-behaved).

Contour integration helps in finding values of integrals which are difficult to solve with standard real analysis methods. It is particularly useful when singularities are present within the contour.

Singularities

In complex analysis, singularities are special points where a function becomes undefined or behaves unusually.

For instance, consider a function \( f(z) = \frac{1}{z^2} \). At \( z = 0 \), the function blows up and is not defined. This is a singularity. Generally:

• A simple pole is a singularity where the function behaves like \( \frac{1}{z-z_0} \) as \( z \to z_0 \).
• A pole of order \( n \) is where it behaves like \( \frac{1}{(z-z_0)^n} \).
• Essential singularities and branch points are types too, but are more complex.

In our exercise, the function had poles at \( z=0 \) of order 2, and \( z=-2 \) of order 1. Recognizing these singularities allows us to use tools like the Cauchy's residue theorem to solve complex integrals.

Residues

Residues in complex analysis refer to a concept closely linked with singularities. They provide a way to summarize the behavior of a function near its singular points and are crucial when using Cauchy's residue theorem.

For a given singularity, the residue can be thought of as the coefficient of \( \frac{1}{z-z_0} \) in the function's Laurent series expansion around the point \( z_0 \). This constant gives us a powerful tool:

• Residues help compute contour integrals.
• The sum of residues inside a contour helps determine the integral's value over that contour.

In simpler terms, if you can find the residues of a function's singular points inside a contour, you can find the integral over that path using Cauchy's residue theorem. In the given problem, residues at the poles \( z=0 \) and \( z=-2 \) are summed up to find the contour integral.