Question

Find the residues of the following functions at the indicated points. Try to select the easiest method. $$ \frac{z+2}{\left(z^{2}+9\right)\left(z^{2}+1\right)} \text { at } z=3 i $$

Short answer

\[ \frac{z+2}{16} \]

Step-by-step solution

Identify the nature of the singularity

Determine the type of singularity at the given point, which is at \(z = 3i\). The given function is \(\frac{z+2}{(z^2+9)(z^2+1)}\). Note that \(z^2 + 9 = (z - 3i)(z + 3i)\), so at \(z = 3i\), \(z^2 + 9 = 0\). Thus, the denominator equals zero, indicating a singularity.

Determine the order of the pole

Examine the terms \((z - 3i)\) in the factor \(z^2 + 9\). Since this factor appears only once, the pole at \(z = 3i\) is a simple pole (first-order pole).

Calculate the residue

For a simple pole at \(z = 3i\), the residue can be found using the formula for the residue at a simple pole: \[ \text{Res}(f, 3i) = \lim_{z \to 3i} (z - 3i) f(z) \] Hence, \[ \text{Res}\left(\frac{z+2}{(z^2+9)(z^2+1)}, 3i\right) = \lim_{z \to 3i} (z - 3i) \frac{z+2}{(z^2+9)(z^2+1)}. \] Simplifying \(f(z)\) at \(z = 3i\), \[ f(z) = \frac{z+2}{(z - 3i)(z + 3i)(z^2 + 1)}, \] Therefore, \[ \text{Res}\left(\frac{z+2}{(z^2+9)(z^2+1)}, 3i\right) = \lim_{z \to 3i} \frac{z+2}{(z + 3i)(z^2+1)} = \frac{3i+2}{(6i)(9-1)} = \frac{3i+2}{48i} = \frac{3i+2}{48i} = \frac{1}{16(3i)} \left( \frac{3i+2}{i} \right) = \frac{3i+2}{48i} = \frac{3i+2}{16} \]

Key concepts

simple pole

In complex analysis, a pole is a type of singularity of a function. A pole occurs when a function goes to infinity at a specific point. There are different orders of poles, such as simple poles, double poles, etc. A simple pole is the specific case where the function behaves like \(\frac{1}{z - a}\) near the point \(a\). This means that the function has a singularity that goes to infinity linearly.

For example, in the given function:

\(\frac{z + 2}{(z^2 + 9)(z^2 + 1)}\),
we found that at \(z = 3i\), the denominator \(z^2 + 9 = 0\), thus causing a singularity. By examining the term \(z - 3i\), and noting that it appears only once, we determine that \(z = 3i\) is a simple pole.

residue calculation

The residue is a key concept in complex analysis, especially in evaluating integrals using the residue theorem. The residue of a function at a point is the coefficient of \(\frac{1}{z - a}\) in the Laurent series expansion of the function around the point \(a\). For a simple pole, there is a straightforward formula to compute the residue. If you have a function \(f(z)\) with a simple pole at \(z = 3i\), the residue can be found using the formula:

\(\text{Res}(f, 3i) = \lim_{z \to 3i} (z - 3i) f(z) \).

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

Simplify the expression inside the limit by substituting the factors. This results in:
\(\text{Res}\bigg(\frac{z + 2}{(z^2 + 9)(z^2 + 1)}, 3i\bigg) = \lim_{z \to 3i} \frac{z + 2}{(z + 3i)(z^2 + 1)}\).

Continuing the simplification leads to our final result: \(\frac{3i + 2}{48i} = \frac{1}{16(3i)} \text{and}\ \frac{3i + 2}{48i} = \frac{1}{16} \text{(3i)}\).

complex function

A complex function is a function that maps complex numbers to complex numbers. These functions are essential in describing physical and mathematical phenomena in higher dimensions. In the specific function given in the solution, \(\frac{z + 2}{(z^2 + 9)(z^2 + 1)}\), the numerator and denominator are both polynomials in terms of \(z\).

The behavior of a complex function near its singularities is crucial for understanding its properties. Singularities can be points where the function does not behave well, such as going to infinity. Understanding and calculating residues help in analyzing these functions and can reveal important features in their behavior.

For this reason, the steps we follow, such as identifying the nature of singularities, determining the order of poles, and calculating residues, are fundamental in complex analysis. By breaking down these steps as shown, it becomes easier to handle and understand even complicated-looking functions.