Question

Find the residues of the following functions at the indicated points. Try to select the easiest method. $$ \frac{\sin ^{2} z}{2 z-\pi} \text { at } z=\frac{\pi}{2} $$

Short answer

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

Step-by-step solution

Identify the Singular Point

The given function is \( \frac{\sin^{2} z}{2z - \pi} \). The singularity is at the point where the denominator equals zero. Set \( 2z - \pi = 0 \) to find this point. Solving for z gives \( z = \frac{\pi}{2} \).

Determine the Type of Singularity

Since the denominator is linear in \( z \) and does not rise to a higher power, and the numerator \( \sin^2 z \) is analytic at \( z = \frac{\pi}{2} \), this point is a simple pole.

Find the Residue at the Simple Pole

The residue of a function \( f(z) = \frac{g(z)}{h(z)} \) at a simple pole \( z = a \) is given by \(\text{Res}(f, a) = \frac{g(a)}{h'(a)} \). Here, \( g(z) = \sin^2 z \) and \( h(z) = 2z - \pi \), then \( h'(z) = 2 \). Evaluate \( g \left( \frac{\pi}{2} \right) \):

Evaluate the Numerator

Evaluate \( \sin^2 \left( \frac{\pi}{2} \right) \). Since \( \sin \left( \frac{\pi}{2} \right) = 1 \), it follows that \( \sin^2 \left( \frac{\pi}{2} \right) = 1 \).

Calculate the Residue

Using the values from previous steps, the residue at \( z = \frac{\pi}{2} \) is \[\text{Res} \left( \frac{\sin^{2} z}{2z - \pi}, \frac{\pi}{2} \right) = \frac{\sin^2 \left( \frac{\pi}{2} \right)}{2} = \frac{1}{2}.\]

Key concepts

Singularity Analysis

In complex analysis, a singularity is a point at which a function is not well-behaved, typically because it involves division by zero. The first step in finding the residue of a function is to identify the singular points (or singularities).
For the given function \(\frac{\text{sin}^2 z}{2z - \text{π}}\), we set the denominator equal to zero and solve for \(z\):\(2z - π = 0\) leads us to the singular point \(z = \frac{π}{2}\). Identifying singularities helps us understand where the function becomes undefined or exhibits unusual behavior.
These points are crucial in various applications, such as calculating integrals and understanding the behavior of complex functions around these points.

Simple Pole

A simple pole is a type of singularity where the function has a zero of order one in the denominator. This means that when the denominator is zero, it does so linearly, i.e., it can be written in the form \((z - a)\) where \(a\) is the point in question.
For the function \(\frac{\text{sin}^2 z}{2z - \text{π}}\), the singularity at \(z = \frac{π}{2}\) is a simple pole because the denominator is a linear term \(2z - π\) and not raised to a higher power.
This implies that the singularity at this point is not any more severe than necessary for a simple pole, simplifying our residue calculation.

Residue Calculation

Calculating residues at simple poles can be straightforward if done methodically. The residue of a function \(f(z) = \frac{g(z)}{h(z)}\) at a simple pole \(z = a\) is given by \(\text{Res}(f, a) = \frac{g(a)}{h'(a)}\).
Let's break it down for our function \(\frac{\text{sin}^2 z}{2z - \text{π}}\):

• Here, \(g(z) = \text{sin}^2 z\)
• and \(h(z) = 2z - π\)
• We find \(h'(z) = 2\).

Next, evaluate \(g(z)\) at \(z = \frac{π}{2}\), and we get \(g \text{(\frac{π}{2})} = \text{sin}^2(\frac{π}{2}) = 1\).
Finally, the residue is calculated by plugging these values into the formula: \(\text{Res} \text{(\frac{\text{sin}^2 z}{2z - \text{π}}, \frac{π}{2})} = \frac{1}{2}\).
Residue calculation greatly simplifies solving complex integrals, particularly in contour integrals in the complex plane.

Complex Functions

Complex functions are mathematical functions involving complex numbers. These functions have properties that are distinctly different from those of real-valued functions.
For instance, the function \(\text{sin}^2 z\) is analytic everywhere in the complex plane, meaning it's differentiable at every point. When combined with rational terms like \(\frac{1}{2z - π}\), the overall function can exhibit different types of singularities based on the nature of the denominator.
Understanding how to manage and manipulate these functions is valuable in fields ranging from engineering to quantum physics. The analysis of these functions, including identifying zeros, poles, and residues, allows us to solve many practical problems more efficiently and effectively.