Question

Find the residues of the following functions at the indicated points. Try to select the easiest method. $$ \frac{1}{(3 z+2)(2-z)} \text { at } z=-\frac{2}{3} \text { and at } z=2 $$

Short answer

Residue at \(z=-\frac{2}{3}\): \(\frac{3}{8}\). Residue at \(z=2\): \(\frac{1}{8}\).

Step-by-step solution

Identify the Poles of the Function

The function given is \(\frac{1}{(3z+2)(2-z)}\). The poles are the values of \(z\) that make the denominator zero. Solving \(3z+2=0\) yields \(z=-\frac{2}{3}\). Solving \(2-z=0\) yields \(z=2\). Therefore, the poles are \(z= -\frac{2}{3}\) and \(z=2\).

Calculate the Residue at \(z=-\frac{2}{3}\)

To find the residue at \(z=-\frac{2}{3}\), write the function as \(\frac{1}{(3z+2)(2-z)}\). The residue at this pole is given by \(\lim_{z \to -\frac{2}{3}} (z + \frac{2}{3}) \frac{1}{(3z+2)(2-z)}\). Simplifying, we get: \[\lim_{z \to -\frac{2}{3}} \frac{1}{2-z} = \frac{1}{2 + \frac{2}{3}} = \frac{1}{\frac{8}{3}} = \frac{3}{8}\]. Therefore, the residue at \(z=-\frac{2}{3}\) is \(\frac{3}{8}\).

Calculate the Residue at \(z=2\)

To find the residue at \(z=2\), write the function as \(\frac{1}{(3z+2)(2-z)}\). The residue at this pole is given by \(\lim_{z \to 2} (z-2) \frac{1}{(3z+2)(2-z)}\). Simplifying, we get: \[\lim_{z \to 2} \frac{1}{3z+2} = \frac{1}{3 \cdot 2 + 2} = \frac{1}{8}\]. Therefore, the residue at \(z=2\) is \(\frac{1}{8}\).

Key concepts

poles in complex functions

In complex analysis, a pole of a function is a type of singularity that behaves like a specific point where the function goes to infinity. Essentially, it's where the denominator of a fraction goes to zero, while the numerator stays finite.
Let's consider the function \(\frac{1}{(3z+2)(2-z)}\).
The poles are found by solving the equations that make the denominator zero. For this function, the poles are at\(-\frac{2}{3}\) and\(2\). When you input these values into the denominator, it becomes zero, indicating the location of the poles.
In simpler terms, poles are points in the complex plane where the function is not defined and tends to infinity. Recognizing these poles is the primary step before calculating residues.

residue calculation

The residue of a function at a given pole is a crucial concept in complex analysis. It helps in evaluating complex integrals using the residue theorem.
Residues can be thought of as the coefficients of the \(\frac{1}{z}\) term in the Laurent series expansion of a function around a given singularity.
To calculate the residue, we use a straightforward formula. For our example function \(\frac{1}{(3z+2)(2-z)}\), the residue at a simple pole \(z = a\) is given by: \[Res(f, a) = \text{lim}_{z \to a} (z - a) f(z)\].[br>This formula essentially 'cancels out' the term that goes to infinity, making it possible to evaluate the function's behavior close to the pole. You'd follow the same formula for any function with simple poles.
In our calculation, we used this method to find the residues at \(-\frac{2}{3}\) and \(2\).

limit method for residues

One of the easiest and most straightforward techniques for finding the residue at a simple pole is the limit method. This involves taking the limit of \( (z - a) f(z) \) as \({z}\) approaches \({a} \), where \(a\) is the pole.
Let's illustrate this with our example. For the given function \(\frac{1}{(3z+2)(2-z)}\):
At \(z = -\frac{2}{3}\), the residue is:
\[ \text{lim}_{z \to -\frac{2}{3}} (z + \frac{2}{3}) \frac{1}{(3z+2)(2-z)} = \frac{3}{8} \]
Similarly, at \(z = 2\), the residue is:
\[ \text{lim}_{z \to 2} (z-2) \frac{1}{(3z+2)(2-z)} = \frac{1}{8} \]
These limits simplify the process, making residue calculation very manageable for students. You just focus on canceling out the problematic term and evaluating what's left using simple algebra.