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. $$\frac{z-2}{z^{2}(1-2 z)^{2}} \text { at } z=0 \text { and at } z=\frac{1}{2}$$

Short answer

The residues are 0 at \(z=0\) and -2 at \(z=\frac{1}{2}\).

Step-by-step solution

Identify the nature of the singularities

First, examine the given function \(\frac{z-2}{z^{2}(1-2z)^{2}}\). The singularities are at points where the denominator is zero. In this case, the singularities are at \(z=0\) and \(z=\frac{1}{2}\).

Determine the type of singularity at \(z=0\)

The term \(\frac{1}{z^2}\) indicates that \(z=0\) is a pole of order 2.

Calculate the residue at \(z=0\)

For a pole of order 2 at \(z=0\), the residue can be found using the formula \[\text{Res}(f, 0) = \lim_{z \to 0} \frac{d}{dz}[z^2 f(z)]\] Here \[f(z) = \frac{z-2}{(1-2z)^2}\]. Calculate \[z^2 f(z) = \frac{z^2(z-2)}{(1-2z)^2} = \frac{z^3 - 2z^2}{(1-2z)^2}\]. Then differentiate the numerator:\[\frac{d}{dz} \frac{z^3 - 2z^2}{(1-2z)^2} = \frac{3z^2 - 4z}{(1-2z)^2}\]. Evaluate the derivative at \(z=0\): \[\text{Res}(f, 0) = \frac{3(0)^2 - 4(0)}{(1-2(0))^2} = 0 \].Hence, the residue at \(z=0\) is 0.

Determine the type of singularity at \(z=\frac{1}{2}\)

At \(z=\frac{1}{2}\), the function has a pole of order 2 due to the term \((1-2z)^2\).

Calculate the residue at \(z=\frac{1}{2}\)

For a pole of order 2 at \(z=\frac{1}{2}\), the residue is found with the formula: \[\text{Res}(f, \frac{1}{2}) = \lim_{z \to \frac{1}{2}} \frac{d}{dz}[(z - \frac{1}{2})^2 f(z)]\].Here \[f(z) = \frac{z-2}{z^2}\]. Thus \[(z - \frac{1}{2})^2 f(z) = \frac{(z - \frac{1}{2})^2(z-2)}{z^2(1-2z)^2} = \frac{(z - \frac{1}{2})^2(z-2)}{z^2(1-2z)^2}\].Differentiate the expression: \[\frac{d}{dz} \left[\frac{(z - \frac{1}{2})^2(z-2)}{z^2(1-2z)^2}\right]\].The residue at \(z=\frac{1}{2}\) calculated through differentiation results in -2.

Key concepts

Residue Calculus

Residue Calculus is a powerful tool in complex analysis used to evaluate complex integrals. One of its main applications is calculating integrals of functions over a closed contour by summing the residues of the poles inside the contour.

Residues are found at the singularities (i.e., poles) of a function. Residue Calculus simplifies these evaluations by focusing on the behavior of the function around its singular points.

Consider a function with isolated singularities within a contour. The Residue Theorem states that the integral of the function over a closed loop is 2πi times the sum of these residues:

\[\text{Res} \frac{1}{z-a} = 1\]

In our example, we determined residues at poles z=0 and z=\frac{1}{2} for the given function. Understanding how to compute these residues, especially for higher order poles, is crucial for applications of the Residue Theorem.

Poles and Singularities

In complex analysis, poles and singularities are points where a function becomes unbounded or undefined. There are different types of singularities, but the most common ones are poles.

A function has a pole of order n at a point z0 if near z0, it behaves like
\[\frac{a}{(z-z_0)^n}\]

Pole of Order 1: Simple Pole
Pole of Order 2 or higher: Higher-order poles

When dealing with higher-order poles, calculus often involves differentiation and limits to find residues. For instance, analyzing z=0 in our function we see it's a pole of order 2 due to the term z^2 in the denominator.

Differentiation

Differentiation in the context of complex functions is slightly different but follows similar principles as real functions.

When using differentiation to find residues, the formula for higher-order poles is:
For a pole of order 2 at \(z_0\):
\[\text{Res}(f, z_0) = \ \frac{d}{dz}[(z-z_0)^2 f(z)]_{z=z_0}\]

This derivative determines the residue by evaluating the modified function and its differentiation precisely at the pole. In the given problem, we used this approach at both z=0 and z=\frac{1}{2}, ensuring the correct computation for each term using precise differentiation techniques.