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. $$\frac{1}{(3 z+2)(2-z)} \text { at } z=-\frac{2}{3} ; \text { at } z=2$$

Short answer

The residue at \(z=-\frac{2}{3}\) is \(-\frac{1}{5}\), and the residue at \(z=2\) is \(-\frac{1}{5}\).

Step-by-step solution

Identify the singularities

The function \(\frac{1}{(3z+2)(2-z)}\) has singularities where the denominator is zero. Solve for \(3z+2=0\) and \(2-z=0\).

Solve for z-values causing singularities

Solve \(3z+2=0\) to get \(z=-\frac{2}{3}\). Solve \(2-z=0\) to get \(z=2\). These are the singular points.

Use the Residue Formula for Simple Poles

For a simple pole at \(z_0\), the residue of a function \(f(z)\) is given by \(\text{Res}(f,z_0) = \frac{g(z_0)}{h'(z_0)}\) where \(f(z)=\frac{g(z)}{h(z)}\).

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

Here, \(f(z) = \frac{1}{(3z+2)(2-z)}\), so \(g(z) = 1\) and \(h(z) = (3z+2)(2-z)\). Factor \(h(z)\) into \(h(z) = (3z+2)(2-z)\) and find that \(h'(z) = 3(2-z) - (3z+2) = -5 \). Thus, \(\text{Res}(f, -\frac{2}{3}) = \frac{1}{-5} = -\frac{1}{5}\).

Find the Residue at \(z=2\)

Here, we use the same function. With \(f(z) = \frac{1}{(3z+2)(2-z)}\), \(g(z) = 1\), and \(h(z) = (3z+2)(2-z)\), we find \(h'(z) = 3(2-z) - (3z+2) = -5\). So, \(\text{Res}(f, 2) = \frac{1}{-5} = -\frac{1}{5}\).

Key concepts

Residue Theorem

The residue theorem is a powerful tool in complex analysis.
It allows us to evaluate complex integrals using residues of singular points.
The theorem states that if you have a complex function \(f(z)\) with isolated singularities within a closed curve \(C\), then the integral of \(f(z)\) around \(C\) is given by:
\[ \oint_C f(z) \, dz = 2 \pi i \sum \text{Res}(f, z_k) \]
Here, \(\sum \text{Res}(f, z_k)\) is the sum of residues of the function at the singular points inside the curve.

This theorem is especially useful for functions with simple poles, as it simplifies the integral computation significantly.

Complex Analysis

Complex analysis studies functions of complex variables.
It delves into differentiability, integration, and series expansion in the complex plane.
Important concepts in complex analysis include:

• Analytic functions: Functions that are differentiable everywhere in their domain.
• Contour integration: Evaluating integrals along paths in the complex plane.
• Singularities: Points where a function fails to be analytic, such as poles and essential singularities.

These ideas find numerous applications in physics, engineering, and other fields due to their ability to simplify and solve otherwise difficult problems.

Singularities

Singularities are points at which a complex function is not well-behaved.
Understanding singularities is key to mastering complex analysis.
They can be categorized into:

• Removable singularities: Points where a function can be redefined to remove the singularity.
• Poles: Points where a function goes to infinity and can be of different orders (simple, double, etc.).
• Essential singularities: Points where the function exhibits wild behavior. Near these points, the function does not settle into any predictable pattern.

For example, in the given problem, the function \( \frac{1}{(3z+2)(2-z)} \) has simple poles at \(z = -\frac{2}{3}\) and \(z = 2\). Understanding where these singularities are helps us determine the residues and solve the integral more easily.