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. $$\frac{e^{i z}}{9 z^{2}+4} \text { at } z=\frac{2 i}{3}$$

Short answer

The residue at \(z = \frac{2i}{3}\) is \(\frac{e^{-\frac{2}{3}}}{12i}\).

Step-by-step solution

Simplify the Denominator

First, factorize the denominator of the function. The denominator is given by \(9z^2 + 4\). This can be written as:\[9z^2 + 4 = (3z + 2i)(3z - 2i)\].

Identify the Singularities

The singularities of the function occur where the denominator is zero. Solving for \(z\) in the factorized form gives us:\((3z + 2i)(3z - 2i) = 0\).Therefore, the singularities are at \(z = \frac{2i}{3}\) and \(z = -\frac{2i}{3}\).

Determine the Residue at the Given Point

Since we are interested in the residue at \(z = \frac{2i}{3}\), we will use the formula for the residue at a simple pole \(z = z_0\):\[\text{Res}\bigg(f, \frac{2i}{3}\bigg) = \frac{g(z_0)}{h'(z_0)}\],where \(f(z) = \frac{g(z)}{h(z)}\) and \(h(z)\) has a simple zero at \(z_0\).In our case, \(g(z) = e^{iz}\) and \(h(z) = 9z^2 + 4\).

Compute \(h'(z)\)

Find the derivative of the denominator \(h(z) = 9z^2 + 4\). Using differentiation rules, we get:\[h'(z) = 18z\].

Evaluate the Residue

Using the formula for the residue at a simple pole, we substitute \(z = \frac{2i}{3}\) into the formulas for \(g(z)\) and \(h'(z)\):\[g\bigg(\frac{2i}{3}\bigg) = e^{i \cdot \frac{2i}{3}} = e^{-\frac{2}{3}}\],\[h'\bigg(\frac{2i}{3}\bigg) = 18 \cdot \frac{2i}{3} = 12i\].Therefore, the residue is:\[\text{Res}\bigg(f, \frac{2i}{3}\bigg) = \frac{e^{-\frac{2}{3}}}{12i}\].

Key concepts

simple pole

In complex analysis, a simple pole is a type of singularity that occurs when a function behaves like \(\frac{c}{z - a}\) near some point \(_{z = a}\). This means, at point \(_{z = a}\), the function has a term that approaches infinity.
In simpler terms, if you can express your function \(f(z) \) as \(g(z)/(z-z_0)\) where \(g(z_0)\ \eq 0\) and \(_{z = z_0}\) is your singularity, then \(z_0\) is a simple pole.
To determine whether a singularity is a simple pole, factorize the denominator of your function and identify the points where it equals zero. If around these points, the numerator is non-zero and finite, those points are simple poles.

residue calculation

Residue calculation is crucial in evaluating complex integrals, particularly in the context of evaluating contour integrals using the residue theorem.
In a nutshell, the residue at a simple pole \(z_0\) is calculated by the formula: \(\text{Res}(f, z_0) = \frac{g(z_0)}{h'(z_0)}\). Here, \(g(z)\) is a function where \(g(z_0)\) is finite and non-zero, and \(h(z)\) is a function with a simple zero at \(z_0\).
Simply put, the residue is the value of the fraction formed by the numerator evaluated at \(_{z = z_0}\) and the derivative of the denominator evaluated at \(_{z = z_0}\).
In our example, we found the residue of the function \(f(z) = \frac{e^{iz}}{9z^2 + 4}\) at \(z = \frac{2i}{3}\). The steps involved simplifying the given function, analyzing the singularities, and substituting into the residue formula. By identifying that \(g(z) = e^{iz}\) and \(h(z) = 9z^2 + 4\), we used differentiation to find \(h(z)\) and then completed the calculation.

singularities

Singularities in complex analysis are points at which a given function is not defined or does not behave in a regular manner. These points can be classified into different types, including poles, essential singularities, and removable singularities.
In our problem, we specifically dealt with simple poles.
To find the singularities of a function, you look for values of \(z\) where the denominator of your fraction equals zero. For instance, in the function \(f(z) = \frac{e^{iz}}{9z^2 + 4}\), the singularities are found by solving \(9z^2 + 4 = 0\).
Once factorized \(9z^2 + 4\) gives us \((3z + 2i)(3z - 2i) = 0\), giving us singularities at \(z = \frac{2i}{3}\) and \(z = -\frac{2i}{3}\). Knowing the locations and types of singularities can help you better understand the behavior of the function and evaluate integrals efficiently.