Problem 36 - A First Course in Complex Analysis with Applications

Question

Discuss how to evaluate \(\oint_{C} e^{1 / z} \sin \left(\frac{1}{z}\right) d z\), where \(C\) is \(|z|=1\). Carry out your ideas.

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Integral

The given problem is a contour integral along the path defined by the curve \( C \), which is the unit circle \(|z| = 1\). The function inside the integral is \( e^{1/z} \sin (1/z) \), which is a complex function. Our task is to evaluate this integral over the specified contour.

Step 2: Analyze Singularity

The function \( e^{1/z} \sin(1/z) \) has a singularity at \( z = 0 \) because both \( 1/z \) and \( \sin(1/z) \) become undefined there. This is an essential singularity, meaning that the Laurent series expansion of the function around this point has infinitely many non-zero negative power terms.

Step 3: Use Residue Theorem

The residue theorem provides a way to evaluate contour integrals by relating them to the sum of residues of the function inside the contour. Since our contour \( C \) is the unit circle \(|z| = 1\), the point \( z = 0 \) lies inside it. We need to find the residue of the given function at \( z = 0 \).

Step 4: Expand the Function as Series

To find the residue at \( z = 0 \), expand \( e^{1/z} \) and \( \sin(1/z) \) as their respective Taylor series: \( e^{1/z} = 1 + (1/z) + (1/2!)(1/z)^2 + \ldots \) and \( \sin(1/z) = (1/z) - (1/3!)(1/z)^3 + \ldots \). Multiply these series expansions to get the series for the entire function.

Step 5: Identify the Residue

The residue is the coefficient of the term \( 1/z \) in the Laurent expansion. After multiplying the series expansions, we find the term \( 1/z \) in the combined series. The product will include terms like \((1/z)\) from \( \sin(1/z)\) and \(1\) from \( e^{1/z}\), giving a residue of \(1\) at \( z = 0 \).

Step 6: Apply the Residue Theorem

Using the residue theorem, the integral \( \oint_{C} e^{1/z} \sin(1/z) \, dz \) is \( 2 \pi i \times \) (residue at \( z = 0 \)), which simplifies to \( 2 \pi i \cdot 1 = 2 \pi i \).

Key Concepts

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis, providing a way to evaluate contour integrals quickly. It is especially useful when dealing with functions that have singularities inside the integration path. In essence, it relates the contour integral of a function around a closed contour to the sum of residues of the function's singular points within that contour.

Here’s how it works: when you have a function that is analytic, except for a few isolated singularities inside a closed contour, the integral of the function over the contour is equal to \[ 2\pi i \sum \text{Residues} \]at those singularities. This simplifies calculating integrals by reducing them to the problem of finding residues.

• Identify singularities within the contour.
• Find residues at these singularities.
• Apply the theorem to integrate over the contour.

Understanding this theorem is crucial for working with complex integrals efficiently, as it simplifies complex calculations.

Laurent Series

The Laurent Series is a representation of a complex function that expresses the function as a series, including terms of both positive and negative powers. This series expansion is especially helpful when studying functions with singularities since these are points where a function is not defined in a typical sense.

Every function that is analytic in an annular region can be represented as a Laurent Series, making it a valuable tool for insights into the behavior of functions near their singularities. The series takes the form:
\[f(z) = \sum_{n=-\infty}^{\infty} a_n (z-c)^n\]

• Negative powers (in the series) indicate singular terms.
• Positive powers represent the function behaving normally as a Taylor series around a point.

Through the Laurent Series, you can identify residues needed for applying the Residue Theorem. Specifically, the residue at a point is the coefficient of the \(1/z\) term, which is crucial for evaluating contours in problems like the one in the exercise.

Singularities

In complex analysis, singularities are points where a function ceases to be analytic, meaning the function doesn’t behave normally or isn’t defined. They are vital to understand because they can significantly impact the outcome of contour integrals.

Different types of singularities behave differently:

• Removable Singularity: The function can be redefined to be analytic. It’s as if the singularity doesn’t exist.
• Pole: At this point, the function behaves like a negative integer power. E.g., \(1/z\)
• Essential Singularity: More complex, as with the function \(e^{1/z} \sin(1/z)\), where the function behaves unpredictably.

In the exercise problem, the function \(e^{1/z} \sin(1/z)\) has an essential singularity at \(z=0\). Understanding this helps in identifying that the Laurent series involves infinitely many negative terms, which influences solving the integral.

Complex Function Analysis

Complex Function Analysis explores the properties and behaviors of functions that take complex numbers as inputs and yield complex numbers as outputs. This field extends calculus into the complex plane, revealing surprising and powerful results.

Key concepts include:

• Analytic Functions: Functions that are differentiable at every point in their domain.
• Contours and Paths: Closed curves in the complex plane over which integration is performed.
• Conformal Mapping: Transformations that preserve angles and are useful in many complex analysis applications.

By utilizing concepts from complex function analysis like singularities and the residue theorem, problems such as evaluating contour integrals become manageable. In our integral problem, complex analysis helps decode the behavior and significance of functions’ singular points, aiding in their integration over specific contours.