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. $$\frac{e^{3 z}-3 z-1}{z^{4}} \text { at } z=0$$

Short answer

The residue at \(z = 0\) is 4.5.

Step-by-step solution

- Identify the Type of Singularity

The function given is \(\frac{e^{3z} - 3z - 1}{z^4}\). We need to determine the type of singularity at \(z = 0\). Observe that the numerator is a regular function, and the denominator is \(z^4\), leading to the point \(z=0\) being a pole of order 4.

- Expand Numerator into a Taylor Series

Next, expand the function \(f(z) = e^{3z} - 3z - 1\) into a Taylor series around \(z = 0\). The Taylor series of \(e^{3z}\) is \(\begin{aligned} e^{3z} &= \sum_{n=0}^{\infty} \frac{(3z)^n}{n!} \ = 1 + 3z + \frac{(3z)^2}{2!} + \frac{(3z)^3}{3!} + \frac{(3z)^4}{4!} + \ldots \end{aligned}\). So, \(f(z) = 1 + 3z + \frac{(3z)^2}{2!} + \frac{(3z)^3}{3!} + \frac{(3z)^4}{4!} + \ldots - 3z - 1 = \frac{(3z)^2}{2!} + \frac{(3z)^3}{3!} + \frac{(3z)^4}{4!} + \ldots \).

- Divide by \(z^4\)

Substitute the expanded series of the numerator into the function \(\frac{e^{3 z} - 3 z - 1}{z^4} = \frac{\frac{(3z)^2}{2!} + \frac{(3z)^3}{3!} + \frac{(3z)^4}{4!} + \ldots}{z^4} = \frac{(3z)^2}{2!z^4} + \frac{(3z)^3}{3!z^4} + \frac{(3z)^4}{4!z^4} + \ldots = \frac{9}{2z^2} + \frac{27}{6z} + \frac{81}{24} + \frac{3^4z}{120} + \ldots \).

- Identify the Residue

To find the residue of \(f(z)\) at \(z = 0\), look for the coefficient of the term \(\frac{1}{z}\) in the Laurent series expansion of the function. From the series \( \frac{9}{2z^2} + \frac{27}{6z} + \frac{81}{24} + \frac{3^4z}{120} + \ldots \), the term \(\frac{27}{6z}\) shows that the residue is \(\frac{27}{6} = 4.5 \).

Key concepts

Residue Calculation

In complex analysis, calculating residues is an essential task. Residues play a key role in the calculation of integrals, especially for functions with poles. The residue of a function at a given point is the coefficient of \(\frac{1}{z - z_0}\) in its Laurent series expansion around that point.
To find residues efficiently:

• Identify the type of singularity.
• Expand the function into a Laurent series.
• Identify the coefficient of \(\frac{1}{z - z_0}\).

To improve understanding, let's revisit an example function \(\frac{e^{3z} - 3z - 1}{z^4}\) at \(z = 0\). We expanded the series and identified the residue as 4.5. Important note: Always ensure the singularity is correctly identified as a removable singularity, a pole, or an essential singularity before proceeding.
This way, you can avoid errors and follow the appropriate method for residue calculation.

Laurent Series

The Laurent series is a more general form of the Taylor series and is particularly useful for representing functions that have singularities. Unlike the Taylor series, the Laurent series can include negative powers of \(z\).
Here’s a refresher on forming a Laurent series:

• Determine the function you need to expand.
• Identify the point around which to expand, especially near the singularity.
• Express the function in terms of powers of \(z\), including negative powers if necessary.

In the example \(\frac{e^{3z}-3z-1}{z^4}\), the Laurent series around \(z=0\) was \(\frac{9}{2z^2} + \frac{27}{6z} + \frac{81}{24} + \frac{81z}{120} + \text{other higher powers}\). Note how both positive and negative powers are present, making the Laurent series a flexible tool for analysis.

Poles and Singularities

Understanding singularities is crucial in complex analysis. A singularity is a point where a function is not defined or behaves 'badly' in some sense. There are different types of singularities:

• Removable Singularity: The function can be redefined to make it analytic.
• Pole: The function goes to infinity in some manner. The order of the pole tells us how severe it is. For example, \(z = 0\) in \(\frac{e^{3z} - 3z - 1}{z^4}\) is a pole of order 4.
• Essential Singularity: The function exhibits chaotic behavior. No redefinition or simple concept like a pole applies.

To proceed with poles, you can often use series expansions or limit definitions to analyze them, as we've seen.

Taylor Series Expansion

The Taylor series expansion is a powerful method to express functions as infinite sums of polynomials. It is fundamental in analytic functions and is often the first step in more complex manipulations.
To form a Taylor series around \(z = 0\):

• Ensure the function is analytic at the point of expansion.
• Compute derivatives of the function at \(z = 0\).
• Use the formula: \(f(z) = \frac{f(0)}{0!} + \frac{f'(0)}{1!} z + \frac{f''(0)}{2!} z^2 + \frac{f'''(0)}{3!} z^3 + \text{higher-order terms}\).

In our example, \(e^{3z}\) was expanded into its Taylor series as it is analytic. This allowed us to later form a Laurent series for \(\frac{e^{3z} - 3z - 1}{z^4}\). Remember, understanding the concept of Taylor series is pivotal in venturing into the deeper waters of complex analysis.