Question

Find the residues of the following functions at the indicated points. Try to select the easiest of the methods outlined above. Check your results by computer. $$\frac{e^{2 z}}{1+e^{z}} \text { at } z=i \pi$$

Short answer

The residue of \( \frac{e^{2z}}{1+e^z} \) at \( z=i \pi \) is \ -1. \

Step-by-step solution

Understand the Problem

To find the residue of the function \( \frac{e^{2z}}{1+e^z} \) at the point \( z = i \pi \,\) the easiest method usually involves identifying the nature of the singularity and using the Laurent series expansion or the residue formula for simple poles.

Check the Nature of the Singularity

Substitute \( z = i \pi \) into the denominator \( 1 + e^z \): \( 1 + e^{i \pi} = 1 + (-1) = 0 \. \) Thus, \( z = i \pi \) is a simple pole because \ 1 + e^z \ has a simple zero at \( z = i \pi \).

Use the Formula for Residue at a Simple Pole

For a function \(\frac{f(z)}{g(z)} \) with a simple pole at \( z = z_0 \, \), the residue is given by \ \text{Res}(f/g, z_0) = \frac{f(z_0)}{g'(z_0)}. \ For \ f(z) = e^{2z} \ and \ g(z) = 1 + e^z \, compute \ f(i \pi) = e^{2i \pi} = 1 \ and \ g'(z) = e^z \ so that \ g'(i \pi) = e^{i \pi} = -1. \

Compute the Residue

Using the formula: \[ \text{Res}\left(\frac{e^{2z}}{1+e^z}, i \pi\right) = \frac{e^{2i \pi}}{e^{i \pi}} = \frac{1}{-1} = -1\]

Verify the Result by Computer

It's always a good practice to check manual mathematical results with computational tools such as Python, MATLAB, or online calculators capable of complex analysis.

Key concepts

Laurent series

To fully grasp residue calculus, understanding the Laurent series is essential. The Laurent series is similar to the Taylor series, but it also accounts for terms with negative powers. It's particularly useful for representing functions with singularities, such as poles. For a function \( f(z) \), the Laurent series around a point \( z_0 \) is:

• \[ f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n \]

What makes the Laurent series special is how it handles the 'bad behavior' of functions near singular points. Unlike the Taylor series, the Laurent series includes coefficients \( a_{-n} \) for negative \( n \). This allows it to accurately capture the behavior of a function near points where it might otherwise seem undefined. If the series only has terms with negative powers, it's termed a `principal part,' helping to specifically analyze singularities like poles.

simple pole

A simple pole, or a pole of order one, is a point where a function goes to infinity in a predictable way. More formally, for a function \[ f(z) = \frac{h(z)}{(z - z_0)} \], \( z_0 \) is a simple pole if \( h(z_0) \eq 0 \). This means the function can be rewritten such that the problematic behavior near \( z_0 \) is isolated as a \( \frac{1}{z - z_0} \) term. In our exercise, the function \[ \frac{e^{2z}}{1+e^z} \] has a simple pole at \( z = i \pi \). When we substitute \( z = i \pi \), the denominator becomes zero: \[ 1 + e^{i \pi} = 0 \], indicating a simple zero for \( 1 + e^z \). Thus, the singularity at \( z = i \pi \) is indeed a simple pole.

residue formula

The residue formula is pivotal in residue calculus. For a function \[ f(z) = \frac{g(z)}{h(z)} \] with a simple pole at \( z_0 \), the residue \text{Res}(f, z_0) \ is calculated by: \[ \text {Res}(f, \ z_0) = \frac{g(z_0)}{h'(z_0)} \]
In our example, for \ f(z) = \frac{e^{2z}}{1+e^z} \, we identify \ g(z) = e^{2z} \ and \ h(z) = 1 + e^z \, with a simple pole at \ z = i \pi \. Substituting in, we compute:

• \ g(i \pi) = e^{2i \pi} = 1 \
• \ h'(z) = e^z \ so \ h'(i \pi) = e^{i \pi} = -1 \

Applying the residue formula, we get: \[ \text{Res} \left\( \frac{e^{2z}}{1+e^z} , i \pi \right\) = \frac{1}{-1} = -1 \] This confirms the residue at \ z = i \pi \ is \ -1 \. To double-check, using computational tools can verify the manual calculation, ensuring accuracy and understanding.