Question

Find the residues of the following functions at the indicated points. Try to select the easiest method. $$ \frac{e^{2 t}}{4 \cosh z-5} \text { at } z=\ln 2 $$

Short answer

The residue is \( \frac{e^{2 t}}{3} \)

Step-by-step solution

- Write the function in terms of variables

The given function is \( \frac{e^{2 t}}{4 \cosh z - 5} \). We need to find the residue at \( z = \ln 2 \). First, express the function in a more manageable form.

- Simplify the denominator

Recall that \( \cosh z = \frac{e^z + e^{-z}}{2} \). Substitute this to get \( 4 \cosh z - 5 = 4 \left(\frac{e^z + e^{-z}}{2}\right) - 5 = 2(e^z + e^{-z}) - 5 \).

- Identify the point \( z = \ln 2 \)

At \( z = \ln 2 \), find \( e^{z} = e^{\text{ln} 2} = 2 \) and \( e^{-z} = e^{-\text{ln} 2} = \frac{1}{2} \).

- Evaluate the denominator at \( z = \ln 2 \)

Substitute these values into the denominator: \( 4 \cosh(\text{ln} 2) - 5 = 2(2 + \frac{1}{2}) - 5 = 2(2.5) - 5 = 5 - 5 = 0 \). Thus, \( z = \ln 2 \) is a singularity.

- Use the residue formula for simple poles

For a simple pole, the residue is given by the limit \( \text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z) \). Here, \( z_0 = \ln 2 \). Let \( f(z) = \frac{e^{2 t}}{4 \, \cosh z - 5} \).

- Compute the limit

Multiply and simplify: \( \lim_{z \to \ln 2} (z - \ln 2) \frac{e^{2 t}}{2(e^z + e^{-z}) - 5} \). The numerator is constant \( e^{2 t} \).

- Derivative of the denominator

Find the derivative of the denominator at \( z = \ln 2 \): \( 2(e^z - e^{-z}) \). At \( z=\ln 2 \), this evaluates to \( 2(2 - \frac{1}{2}) = 2 \cdot 1.5 = 3 \).

- Final computation

Using the residue formula, the residue at \( z = \ln 2 \) is \( \frac{e^{2 t}}{3} \).

Key concepts

Complex functions

Complex functions are fundamental elements in complex analysis. They take complex numbers as inputs and produce complex numbers as outputs.
Understanding complex functions involves familiarizing yourself with their properties and behavior beyond the real number line.
Consider a function \( f(z) \), where \( z \) is a complex variable. Key operations and concepts arise, such as differentiation and integration, similar to real functions. However, the complexity increases as you introduce imaginary components.

• The domain of a complex function can include all complex numbers or a subset of them.
• Common functions include polynomials, exponential functions, and trigonometric functions modified for the complex plane.
• Complex functions can exhibit rich behaviors, such as intricate symmetry and periodicity.

To analyze and interpret these functions effectively, foundational knowledge in both real analysis and basic complex arithmetic is essential.

Poles and residues

Poles and residues are crucial concepts when diving into complex functions.
A pole is a type of singularity where a function's value grows without bound as it approaches a specific point.
Residues, on the other hand, are values that describe the behavior of integrals around these singularities.
Identifying poles and calculating residues are essential techniques in complex analysis. Here's a breakdown:

• Simple poles: Singularities where the function can be expressed straightforwardly around the point.
• Higher-order poles: Singularities where the function involves higher powers of terms in the denominator.
• Residue: For a given function at a simple pole, the residue can be computed using the formula \( \text{Res}(f, z_0) = \text{lim}_{z \to z_0} (z - z_0)f(z) \).

Understanding these concepts enriches your ability to solve complex integral problems and recognize the intricate behavior of functions around singular points.

Residue theorem

One of the main tools in complex analysis is the Residue Theorem.
This theorem provides a method to evaluate complex integrals using the residues of a function’s poles.
It states that if a function is analytic inside and on a simple closed contour, except for a finite number of singular points, the integral around the contour is \( 2\pi i \) times the sum of the residues at those points.
Mathematically, if \( C \) is a positively oriented, simple closed contour and \( f(z) \) is analytic on and inside \( C \), except for isolated singularities \( z_1, z_2, \.\.\, z_n \), then:
\[ \int\_C f(z) \,dz = 2\pi i \sum\_k \text{Res}(f, z\_k) \]
The main steps in using the Residue Theorem:

• Identify the singularities (poles) within the contour.
• Calculate the residues at these poles.
• Sum the residues and multiply by \( 2\pi i \).

This theorem simplifies solving complex integrals, allowing you to bypass direct integration by focusing on the function’s behavior at its singularities.

Cosh function

The Hyperbolic Cosine function, denoted as \( \cosh(z) \), is an indispensable function in complex analysis.
It is defined as \( \cosh(z) = \frac{e^z + e^{-z}}{2} \).
Understanding its properties is crucial when dealing with complex functions, as seen in our example problem.
Here are the main aspects:

• Even function: \( \cosh(-z) = \cosh(z) \).
• Periodicity: Unlike trigonometric functions, hyperbolic functions are not periodic in the complex plane.
• Relation to exponential functions: Expressing \( \cosh(z) \) using exponentials aids in simplifying and solving complex expressions.

In the given exercise, substituting \( \cosh(z) \)’s definition made it easier to identify the singularity and compute the residue.
These properties make \( \cosh(z) \) a powerful tool in simplifying complex computations and identifying key points in functions.