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

Short answer

The residues are -2 at z=0 and 1 at z=1.

Step-by-step solution

- Understanding the Problem

We need to find the residues of the function \( \frac{z-2}{z(1-z)} \) at two points: \( z = 0 \) and \( z = 1 \). Residues can be found using various methods, but here we'll use the residue theorem for simple poles.

- Identifying the Simple Poles

Examine the function \( \frac{z-2}{z(1-z)} \). The poles of the function are the values of \( z \) that make the denominator zero. Hence, setting \( z = 0 \) and \( 1 - z = 0 \) gives us the poles \( z = 0 \) and \( z = 1 \).

- Residue at \( z = 0 \)

The residue of a function \( f(z) \) at a simple pole \( z=a \) can be calculated as follows: \[ \text{Res}(f, a) = \frac{\text{Numerator}}{ \frac{d}{dz}(\text{Denominator})}\bigg|_{z=a} \] For \( z = 0 \), the numerator is \( z - 2 \) and the denominator is \( z(1-z) \). Calculate the derivative of the denominator: \[ \frac{d}{dz}(z(1-z)) = 1 - 2z \bigg|_{z=0} = 1 \] Thus, the residue at \( z = 0 \) is: \[ \text{Res}\bigg(\frac{z-2}{z(1-z)}, 0\bigg) = \frac{0-2}{1} = -2 \]

- Residue at \( z = 1 \)

Using the same method, for \( z = 1 \), the numerator is \( z - 2 \) and the denominator is \( z(1-z) \). Calculate the derivative of the denominator: \[ \frac{d}{dz}(z(1-z)) = 1 - 2z \bigg|_{z=1} = -1 \] Therefore, the residue at \( z = 1 \) is: \[ \text{Res}\bigg(\frac{z-2}{z(1-z)}, 1\bigg) = \frac{1-2}{-1} = 1 \]

Key concepts

simple poles

In complex analysis, a simple pole refers to a type of singularity of a complex function. Specifically, it is a point where the function goes to infinity in such a way that when you multiply the function by \(z - a\), it becomes a well-defined, non-zero, and finite value. This happens when the order of the pole is one, meaning the function can be expressed as: \[ f(z) = \frac{g(z)}{(z - a)} \] where \(g(z)\) is analytic and non-zero at \(z = a\). Simple poles are important because they are easy to handle and allow for straightforward calculations of residues. Always look for points where the denominator turns zero while analyzing poles.

residue theorem

The residue theorem is a powerful tool in complex analysis. It helps in evaluating complex integrals by focusing on the residues of singularities within an integral's path. A residue at a simple pole \(z = a\) of a function \(f(z)\) is calculated as:

• Find the Laurent series expansion of \(f(z)\) around \(a\).
• Identify the coefficient of \(\frac{1}{z-a}\).

Alternatively, for a function expressed as \(f(z) = \frac{g(z)}{(z - a)h(z)}\) near the pole \(z = a\), we can calculate the residue as:
An immediate application of this is:
\[ \text{Res}(f, a) = \frac{g(a)}{h(a)} \]
For example, in the given problem for the function \( \frac{z - 2}{z(1 - z)} \), we found the residues at \( z = 0 \) and \( z = 1 \).
The residue at \(z = 0\) was:
\[\text{Res}\bigg(\frac{z - 2}{z(1 - z)}, 0\bigg) = -2 \]
and at \(z = 1\) was:
\[\text{Res}\bigg(\frac{z - 2}{z(1 - z)}, 1\bigg) = 1 \].

complex functions

Complex functions take complex numbers as inputs and provide complex outputs. They are more intricate than real functions and are a core part of complex analysis. A function \(f(z)\) is complex if it is composed of both real and imaginary parts. The simplest form of a complex function is given by:

• \[ z = x + iy \] where x and y are real numbers, and i is the imaginary unit.
• A complex function thus can be written as \[ f(z) = u(x, y) + iv(x, y) \]
• where \(u\) and \(v\) are real functions.

These functions must adhere to specific rules, like being analytic, to attain valuable properties. For example, the given function \( \frac{z - 2}{z(1 - z)} \) in the problem is a rational function. It is analytic everywhere except where its denominator is zero. Understanding this helps in identifying the poles and calculating residues accurately.