Problem 34 - A First Course in Complex Analysis with Applications

Question

Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. $$ \oint_{C} e^{4 /(z-2)} d z, C:|z-1|=3 $$

Step-by-Step Solution

Answer

The integral evaluates to \( 2\pi i \).

Step 1: Identify Singularities

To apply Cauchy's residue theorem, identify the singularities of the function within the contour. The function is \( f(z) = e^{4/(z-2)} \). It has a singularity at \( z = 2 \), which lies inside the contour \( C: |z-1|=3 \) since \(|2-1|=1 < 3\).

Step 2: Compute the Residue

Find the residue of \( f(z) \) at \( z = 2 \). Since the function \( e^{rac{4}{z-2}} \) can be expanded as a Laurent series around this point, we need the coefficient of \( \frac{1}{z-2} \). The residue at \( z = 2 \) is simply \( e^{4/(z-2)} \) evaluated at the leading singular term's coefficient, which is 1 when expanded in Laurent series form.

Step 3: Apply Cauchy's Residue Theorem

Cauchy's Residue Theorem states \( \oint_C f(z) \, dz = 2 \pi i \times \sum \text{Residues inside } C \). Here, there is only one residue inside \( C \) at \( z=2 \), and the residue is 1. Thus, \( \oint_{C} e^{4 /(z-2)} \, dz = 2 \pi i \).

Key Concepts

Complex Integration

Complex integration involves integrating a function along a path or contour in the complex plane. These integrals are similar to line integrals in real analysis, but they take into account the complex nature of the functions. To solve such integrals, you need both a path or contour and a complex function. Often, complex functions have what are called singularities, points where they are undefined.

In many cases, we encounter integrals of the form \( \oint_{C} f(z) \, dz \), where \( f(z) \) is a complex function and \( C \) represents a closed contour around the complex plane. The choice of contour can greatly affect the outcome of the integral, especially if it includes singularities of the function. When a singularity is found within the contour, Cauchy's Residue Theorem can be applied to simplify the evaluation.

• A contour must be closed.
• Functions must be analytic on and inside the contour—except at singularities.
• Complex integration tasks often involve evaluating integrals that would be much more difficult in the real domain.

Laurent Series

The Laurent Series is a key tool in complex analysis for representing complex functions. It extends the idea of a Taylor series, but it includes terms that have negative powers, allowing it to represent functions near singularities. This makes it invaluable when dealing with functions that have parts that tend toward infinity, such as bad behavior around poles.

Functions can often be locally expanded using Laurent series, making calculation of residues simple. For instance, if a function has a simplified form of \( \frac{1}{z-a} \), then it has a singularity at \( z=a \), and the residue is the coefficient of \( \frac{1}{z-a} \) in its Laurent expansion.

Key points about Laurent series:

• They can accommodate singularities by including terms with negative exponents.
• A Laurent series converges in an annular region between two radii centered on the singularity.
• Residues, which are crucial for complex integration, are found by locating the term with a \( \frac{1}{z-a} \) coefficient in the Laurent series.

Singularities

In the context of complex functions, singularities are points where a function ceases to be well-defined. They are critical when performing complex integration because they often dictate whether a contour integral can be simplified using techniques like Cauchy's Residue Theorem.

There are different types of singularities, and understanding these can be critical to solving integrals.

• Simple poles: Resemble \( \frac{1}{z-a} \). The function behaves like \( \frac{1}{z} \) around the singular point.
• Essential singularities: Infinite number of terms in both positive and negative powers in the Laurent series. They tend to make integration more complex.
• Removable singularities: Points where the function is undefined, but can be defined so that the function is analytic at the point.

Identifying and understanding these singularities is a crucial step in using Cauchy's Residue Theorem to evaluate integrals over complex functions. In the given problem, the singularity at \( z=2 \) is a simple pole, making it straightforward to find the residue and apply the theorem.