Question

Evaluate the following integrals by computing residues at infinity. Check your answers by computing residues at all the finite poles. (It is understood that \(\oint\) means in the positive direction.) $$\oint \frac{z^{2} d z}{(2 z+1)\left(z^{2}+9\right)} \quad \text { around }|z|=5$$.

Short answer

Integral = 2\pi i \left[ \frac{1}{37} + (-\frac{3}{2i-1}) + (-\frac{3}{2i+1}) \right].

Step-by-step solution

- Identify the poles of the integrand

First, identify the poles of the integrand \(\frac{z^{2} d z}{(2 z+1)\left(z^{2}+9\right)}\).The poles are found by setting the denominator to zero:\(2z + 1 = 0\) and \(z^2 + 9 = 0\).Solving these equations gives:\z = -\frac{1}{2}\ and \(z = \pm 3i\).

- Verify poles inside the contour

The contour is \(|z|=5\), which means we are considering the circle of radius 5 centered at the origin.Check each pole:\|z| = |-\frac{1}{2}| = 0.5 < 5|3i| = 3 < 5|-3i| = 3 < 5All poles are inside the contour.

- Compute residue at \(z = -\frac{1}{2}\)

To compute the residue at \(z = -\frac{1}{2}\), use the formula for simple poles:\Res(f, -\frac{1}{2}) = \lim\_{z \to -\frac{1}{2}} (z + \frac{1}{2}) \frac{z^{2}}{(2 z+1)(z^{2}+9)}This simplifies to: \Res(f, -\frac{1}{2}) = \frac{(-\frac{1}{2})^2}{2(-\frac{1}{2})+1)((-\frac{1}{2})^{2} + 9)} = \frac{1/4}{9.25} = \frac{1}{37}\.

- Compute residue at \(z = 3i\)

To compute the residue at \(z = 3i\), use the simple pole formula:\Res(f, 3i) = \lim\_{z \to 3i} (z - 3i) \frac{z^{2}}{(2 z+1)(z^{2}+9)}This simplifies to: \Res(f, 3i) = \frac{(3i)^{2}}{2(3i)+1)((3i)^{2}+9)} = \frac{-9}{-6i+1} = -\frac{3}{2i - 1}\.

- Compute residue at \(z = -3i\)

To compute the residue at \(z = -3i\), use the simple pole formula:\Res(f, -3i) = \lim\_{z \to -3i} (z + 3i) \frac{z^{2}}{(2 z+1)(z^{2}+9)}This simplifies to: \Res(f, -3i) = \frac{(-3i)^{2}}{2(-3i)+1)((-3i)^{2}+9)} = \frac{-9}{-6i-1} = -\frac{3}{2i+1}\.

- Evaluate the integral using residues

The value of the integral is \(2\pi i\) times the sum of the residues within the contour:\Integral = 2\pi i \left[ \frac{1}{37} + (-\frac{3}{2i-1}) + (-\frac{3}{2i+1}) \right].Simplify the expression to get the final result.

Key concepts

residue theorem

The residue theorem is a powerful tool in complex analysis that helps evaluate complex integrals. It states that if you have a meromorphic function (a complex function that is differentiable except at some isolated points called poles) and a closed contour around these poles, the integral of the function around the contour is equal to \(2\pi i\) times the sum of the residues at those poles.
In our exercise, we are asked to find the integral by computing residues. This involves identifying the poles within a given contour and calculating the residues at these poles. The residue theorem simplifies the process significantly, avoiding the need for direct and often complex integration.
Residues can be thought of as the coefficients of the \(\frac{1}{z - z_0}\) term in the Laurent series expansion of a function around a pole \(z_0\). They encapsulate the behavior of a function near its singularities.

poles

Poles are specific types of singularities where a complex function behaves like \(\frac{1}{(z - z_0)^n}\) near some point \(z_0\). There are different types of poles:

• **Simple poles**: These occur when the function behaves like \(\frac{1}{z - z_0}\) around \(z_0\). For example, \(f(z) = \frac{1}{z - 1}\) has a simple pole at \(z = 1\).
• **Higher-order poles**: These occur when the function behaves like \(\frac{1}{(z - z_0)^n}\) for \(n > 1\). For example, \(f(z) = \frac{1}{(z - 1)^2}\) has a pole of order 2 at \(z = 1\).

In the given exercise, the integrand \(\frac{z^{2} dz}{(2 z+1)(z^{2}+9)}\) has three poles. By setting the denominator to zero, we find that the poles are at \(z = -\frac{1}{2}\), \(z = 3i\), and \(z = -3i\). All of these are simple poles, which makes our residue computation more straightforward.

contour integration

Contour integration involves integrating complex functions along a specified path or contour in the complex plane. This path is usually a closed loop, allowing us to use powerful tools like the residue theorem.
In the current exercise, the contour is specified as \(|z| = 5\), meaning we are considering a circle with radius 5 centered at the origin. This contour encloses all three poles \((-\frac{1}{2}, 3i, -3i)\) of the integrand. When performing contour integration, it's crucial to ensure that your contour encloses all the poles you're interested in, but no more.
To evaluate an integral using contour integration:

• **Identify the poles inside the contour.**
• **Compute the residues at these poles.**
• **Sum these residues.**
• **Multiply the sum by \(2\pi i\) to get the value of the integral.**

complex analysis

Complex analysis is the study of functions that operate on complex numbers. It extends real analysis into the complex plane and introduces new concepts and techniques essential for solving problems involving complex numbers.
Some key concepts in complex analysis include:

• **Analytic functions**: Functions that are differentiable in some domain of the complex plane.
• **Contour integration**: Integrating along a path in the complex plane.
• **Poles and residues**: Special points and their associated coefficients in the Laurent series expansion of a function.

In our exercise, we are applying complex analysis techniques such as identifying poles, computing residues, and using the residue theorem to evaluate a given integral. These techniques are not only helpful in academic exercises but also play a significant role in fields like electrical engineering, quantum physics, and applied mathematics.