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. $$\frac{z}{1-z^{4}} \text { at } z=i$$

Short answer

The residue of \( \frac{z}{1-z^{4}} \) at \(z=i \) is \ \frac{1}{2} \.

Step-by-step solution

- Identify the function and the point

The function to consider is \(\frac{z}{1-z^{4}}\) and the specified point is \(z=i\).

- Expand the denominator

Factor the denominator \(1-z^{4} = (1-z^2)(1+z^2) = (1-z)(1+z)(1-iz)(1+iz)\).

- Simplify at the specified point

At \(z=i\), one of the factors in the denominator is \(1-iz=1-i(i)=1+1=2\). This is non-zero, hence the singularity is due to the other terms.

- Determine the type of singularity

The point \(z=i\) is a simple pole (order 1) of the function because it is zero of \(1+iz\).

- Use the residue formula for a simple pole

The residue at \(z=i\) for a function \(\frac{f(z)}{g(z)}\) where \(f(i)\) is analytic and \(g(z)\) has a simple zero at \(z=i\): \[ \text{Res}\bigg(f(z)=\frac{z}{1-z^4}, z=i\bigg) = \frac{f(i)}{g'(i)} = \frac{i}{(1+z^2)'(i)} = \frac{i}{(2i)} = \frac{1}{2} \]

Key concepts

Simple Pole

A simple pole is a specific type of singularity in complex analysis. It occurs when a function has a denominator with a zero of order one. In other words, it is a point where the function goes to infinity, but not too fast. Consider the function \(\frac{z}{1-z^{4}}\). When looking at the point \(z=i\), we notice that \(1-iz\) goes to zero because it results in \(1-ix=0\). This means \(z=i\) is a simple pole of the function. To identify if it's a simple pole:

• Factor the denominator.
• Check the power of the zero where the function becomes undefined.

Here, \(1+iz\) is the factor making the denominator zero and since it is to the first power, i.e., \(1-iz\), \(z=i\) is indeed a simple pole.

Residue Calculation

Calculating residues at a simple pole is straightforward with the residue formula. The residue of a function at a simple pole provides the coefficient of \(\frac{1}{z-a}\) in its Laurent series expansion. For function \(\frac{z}{1-z^{4}}\) at \(z=i\), we use the formula:

• Identify \(f(z)\) and \(g(z)\) where the function is \(\frac{f(z)}{g(z)}\).
• Determine \(g'(i)\), the derivative of the part making the denominator zero at the specified pole.

Here, \(\frac{z}{1-z^4}\), we let \(f(z)=z\) and \(g(z)=1-z^4\). Since the zero at \(z=i\) is from \(1+iz\), its derivative \(g'(z)=4z^3\) at \(z=i\) gives \(4i^3=4 \times (-i)=-4i\). By substituting this into the residue formula,
Residue \(\text{Res}\bigg(\frac{z}{1-z^4}, i\bigg) = \frac{i}{-4i} = \frac{1}{2}\).

Complex Functions

Complex functions extend real functions to the complex plane, allowing input and output to be complex numbers. Functions such as \(\frac{z}{1-z^4}\) exhibit unique properties, like singularities, which don't appear in the same manner in real functions. Complex functions have special places where they aren't defined—known as poles, essential singularities, or branch points.
Here are standard steps to analyze and work with complex functions:

• Identify points where the function becomes undefined.
• Determine the nature of these singularities (pole, essential, removable).
• Use tools like residue calculus to evaluate integrals around these points.

With \(\frac{z}{1-z^4}\), one examines where \(1-z^4\) is zero, identifies singularities, and studies their behavior to understand the function better. Complex analysis provides rich tools to manage and utilize these properties effectively.