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. $$\frac{z+2}{\left(z^{2}+9\right)\left(z^{2}+1\right)} \text { at } z=3 i$$

Short answer

Residue is \frac{3i+2}{-24(2)}.

Step-by-step solution

Identify the singularity

The function \(\frac{z+2}{(z^2+9)(z^2+1)}\) has singularities where the denominator equals zero. These occur at the points where \(z^2+9=0\) and \(z^2+1=0\). Solving these, we have \(z^2=-9\), giving \(z=\pm 3i\), and \(z^2=-1\), giving \(z=\pm i\). The specific singularity of interest is at \(z=3i\).

Simplify the denominator near the singularity

To find the residue, look at the behavior of the function near \(z=3i\). Rewrite the function by factoring out the term related to \(z-3i\) in the denominator. \[\frac{z+2}{(z-3i)(z+3i)(z-i)(z+i)}\]

Focus on the residue computation

Since \(z=3i\) is a simple pole, the residue at \(z=3i\) can be found by taking the limit: \[\text{Res}\bigg(\frac{z+2}{(z^2+9)(z^2+1)}, 3i\bigg) = \lim_{z\to 3i} (z-3i) \frac{z+2}{(z^2+9)(z^2+1)}\]

Compute the limit

To find the residue, evaluate the limit: \[\text{Res}\bigg(\frac{z+2}{(z^2+9)(z^2+1)}, 3i\bigg)= \lim_{z\to 3i} \frac{(z-3i)(z+2)}{(z-3i)(z+3i)(z-i)(z+i)}= \lim_{z\to 3i} \frac{z+2}{(z+3i)(z-i)(z+i)}\]

Substitute and simplify

Substitute \(z \to 3i\) into the simplified expression: \[\text{Res}\bigg(\frac{z+2}{(z^2+9)(z^2+1)}, 3i \bigg) = \frac{(3i+2)}{(3i+3i)(3i-i)(3i+i)}= \frac{3i+2}{(6i)(2i)(4i)}= \frac{3i+2}{-48i^3} = \frac{3i+2}{-48(-i)}= \frac{3i+2}{-48i}\]

Calculate the residue

Final calculation: \[\text{Res}\bigg(\frac{z+2}{(z^2+9)(z^2+1)}, 3i \bigg) = \frac{\frac{3i+2}{-48i}} = \frac{3i+2}{-48i} = \frac{3i+2}{-\frac{48 2}{i}}\]

Key concepts

Complex Singularities

A complex singularity, in simple terms, is a point where a function fails to be analytic, meaning it cannot be represented by a Taylor series around that point. Singularities are crucial in complex analysis because they often indicate points where a function takes on infinite values or exhibits unusual behavior.

There are different types of singularities:

• **Isolated singularities**: Points where the function is not analytic but analytic everywhere else in some neighborhood around the point.
• **Poles**: A type of isolated singularity where the function goes to infinity.
• **Essential singularities**: Points where the behavior of the function is more erratic than poles.

In our given problem, we are focusing on finding the residue at the singularity where the denominator of the function equals zero, specifically at the point where \(z = 3i\). This makes \(z=3i\) a pole of the function.

Simple Poles

In complex analysis, a simple pole is a type of singularity where a function goes to infinity linearly—that is, like \(1/(z - z_0)\). For a function \(f(z)\), if it has a simple pole at \(z = z_0\), then near \(z_0\), the function can be represented as: \ \f(z) \ = \frac{g(z)}{(z - z_0)} \ , \where \ g(z) \ is analytic and non-zero at \ z = z_0\

Identification and handling of simple poles involve straightforward steps:

• *Looking at the denominator*: Find where the denominator equals zero.
• *Using limits*: To find the residue at a simple pole \(z_0\), multiply the function by \((z - z_0)\) and take the limit as \(z\) approaches \(z_0\).

In our exercise, the given function: \ \f(z) \ = \frac{z+2}{(z^2+9)(z^2+1)} \ has a simple pole at \ z = 3i \ . By using the limit approach:

\ \text{Res}\bigg(\frac{z+2}{(z^2+9)(z^2+1)}, 3i\bigg) = \ \lim_ {z\to 3i} \ (z - 3i) \frac{z + 2}{(z^2 + 9)(z^2 + 1)} \.

Residue Theorem

The Residue Theorem is one of the most powerful tools in complex analysis. It links the residues of a function to the integral of the function around a closed contour. Mathematically, it states: \ If \ f \ is analytic inside and on a simple closed contour \ C \ , except for a finite number of singularities \ z_k \ inside \ C \ , then \ \frac{1}{2\boldsymbol{\text{\(\tau\)}i}i} \oint_{C} f(\boldsymbol{z}) \tex{\boldsymbol{d}z}= \sum \tex{Res(f,z_k)} \

Using this theorem, we can simplify many complex integrals by focusing only on the residues of the singularities inside the contour.
In this problem, understanding the theorem helps in confirming that once we've calculated the residues correctly, it can indeed be used to evaluate integrals around these points.