Question

Find the residues of the given function at all poles. Take \(z=r e^{i \theta}\) \(0 \leq \theta<2 \pi\).$$\frac{\ln z}{(2 z-1)^{2}}$$

Short answer

The residue at the pole \(z = \frac{1}{2}\) is \(\frac{1}{2}\).

Step-by-step solution

- Identify the poles

The poles of the function can be found by analyzing the denominator \((2z-1)^2\). Solving \(2z-1=0\) gives \(z = \frac{1}{2}\). Therefore, the function has a pole at \(z = \frac{1}{2}\).

- Determine the order of the pole

The given pole at \(z = \frac{1}{2}\) is a second-order pole because the denominator is \( (2z-1)^2 \).

- Calculate the residue

To find the residue at a second-order pole \(z_0 = \frac{1}{2}\), use the formula: \[ \text{Res}(f, z_0) = \frac{d}{dz} \bigg( (z - z_0)^2 f(z) \bigg) \bigg|_{z = z_0} \]. Rewrite the function as \((z - \frac{1}{2})^2 \frac{\text{ln}(z)}{(2z-1)^2} = \frac{\text{ln}(z)}{4} \). Now differentiate \(\frac{\text{ln}(z)}{4} \) with respect to \(z\): \(\frac{d}{dz} \bigg( \frac{\text{ln}(z)}{4} \bigg) = \frac{d}{dz} \bigg( \frac{\text{ln}(z)}{4} \bigg) = \frac{1}{4z}\). Evaluate this derivative at \(z = \frac{1}{2}\): \[ \frac{1}{4 \frac{1}{2}} = \frac{1}{2} \]. So, the residue at \(z = \frac{1}{2}\) is \(\frac{1}{2}\).

Key concepts

Residues

In complex analysis, residues are a key concept used in many computations involving complex functions. A residue is the coefficient of \frac{1}{z-z_0} in the Laurent series expansion of a complex function around a singularity\ (pole)\, where\ \(z_0\) is the location of the singularity. The residue can provide valuable information about the behavior of the function near that singularity and is essential in evaluating complex integrals using the residue theorem.

• To find the residue, identify and expand the function into its Laurent series near the pole.
• The residue is the coefficient of the \frac1{z-z_0} \ term.
• For higher-order poles, special techniques (like differentiation) are used.

Poles

Poles are points where a complex function becomes unbounded. They are specific types of singularities that are central to the analysis of complex functions. Identifying the poles of a function involves analyzing its denominator for values that make the function undefined or infinite. Types of poles:

• Simple Poles: These are poles of order one, where the function grows without bound as \ \(z \rightarrow z_0 \). The term \ (\frac1{z-z_0}) \ appears in the Laurent series.
• Higher-order Poles: These occur when the function has a more rapid divergence than a simple pole. For example, a pole of order two has a Laurent series containing \ (\frac1{(z-z_0)^2}) \.

Second-Order Pole

A second-order pole is a singularity where the function's denominator has a quadratic term like \ (2z-1)^2 \ in our given problem. The steps to find the residue at a second-order pole are more intricate than for a simple pole.

• Identify the Pole: Solve the denominator \ (2z-1)^2 = 0 \ to find \ \(z = \frac1{2}\)
• Determine the Order: Recognize that the term \ (2z-1)^2 \ signifies a second-order pole.
• Calculate the Residue: Use the residue formula for second-order poles: \ \text{Res}(f, z_0) = \frac{d}{dz} \bigg( (z - z_0)^2 f(z) \bigg) \bigg|_{z = z_0} \.

Complex Functions

Complex functions form the backbone of complex analysis, an area of mathematics that deals with functions of a complex variable. These functions have a rich structure and exhibit fascinating properties not visible in real functions. Key Points to Understand Complex Functions:

• Analytic Functions: These are functions that are differentiable at every point in their domain. Differentiability implies that the function has a Taylor series expansion at each point.
• Singularities: Points where the function fails to be analytic. These include poles, branch points, and essential singularities.
• Laurent Series: This is a series expansion for functions near singularities, containing negative powers of \ \(z\).
• Residue Theorem: A powerful tool for evaluating integrals, it relates the sum of residues of a function inside a contour to the integral around that contour.