Question

Find the residues of the following functions at the indicated points. Try to select the easiest method. $$ \frac{\cosh z-1}{z^{7}} \text { at } z=0 $$

Short answer

The residue of \( \frac{\text{cosh}(z) - 1}{z^7} \) at \( z = 0 \) is \( \frac{1}{720} \).

Step-by-step solution

Understand the Function and Residue

The function to analyze is \( f(z) = \frac{\text{cosh}(z) - 1}{z^7} \). We need to find the residue at the point \( z = 0 \). Residues are associated with terms in the Laurent series expansion of the function, particularly those with a \( \frac{1}{z} \) term.

Expand the Numerator using Taylor Series

Expand \( \text{cosh}(z) \) using its Taylor series: \( \text{cosh}(z) = 1 + \frac{z^2}{2!} + \frac{z^4}{4!} + \frac{z^6}{6!} + \frac{z^8}{8!} + \frac{z^{10}}{10!} + \text{higher order terms} \). Then, \( \text{cosh}(z) - 1 = \frac{z^2}{2!} + \frac{z^4}{4!} + \frac{z^6}{6!} + \frac{z^8}{8!} + \frac{z^{10}}{10!} + \text{higher order terms} \).

Form the Laurent Series

We can now express the function \( f(z) \) as: \[ f(z) = \frac{\text{cosh}(z) - 1}{z^7} = \frac{\frac{z^2}{2!} + \frac{z^4}{4!} + \frac{z^6}{6!} + \text{higher order terms}}{z^7} \]. Simplify this to get\[ f(z) = \frac{z^2}{2!z^7} + \frac{z^4}{4!z^7} + \frac{z^6}{6!z^7} + \frac{z^8}{8!z^7} + \frac{z^{10}}{10!z^7} + \text{higher order terms} \]This simplifies to: \[ f(z) = \frac{1}{2!z^5} + \frac{1}{4!z^3} + \frac{1}{6!z} + \frac{z}{8!} + \frac{z^3}{10!} + \text{higher order terms} \]

Identify the Term with \( 1/z \)

To find the residue, we need the coefficient of the \( \frac{1}{z} \) term. From the simplified Laurent series, we see that the \( \frac{1}{z} \) term is \( \frac{1}{720z} \), where \( 720 \) is \( 6! \).

State the Residue

The residue of the function at \( z = 0 \) is the coefficient of the \( \frac{1}{z} \) term, which is \( \frac{1}{720} \).

Key concepts

Laurent Series Expansion

In complex analysis, a Laurent series is a representation of a complex function as a series which can include terms of both positive and negative powers of the variable. This is especially useful for functions with singularities (points where the function isn’t defined). It's like extending a Taylor series, but it can handle more complicated scenarios where the function behaves badly at certain points.
When we encounter a function with a singularity, we usually want to express it as a Laurent series so we can understand the behavior near that point. The general form of a Laurent series around a singularity at point c is:
\[ f(z) = \frac{a_{-n}}{(z-c)^n} + \frac{a_{-(n-1)}}{(z-c)^{(n-1)}} + ... + a_0 + a_1(z-c) + a_2(z-c)^2 + ... \]
This series includes both negative and non-negative powers of \(z - c\). Specifically, negative powers are key to capturing the behavior near a singularity.
In our example, we expanded \(\text{cosh}(z) - 1\) and divided by \(z^7\) to get its Laurent series. This helped us identify the important terms for calculating residues.

Taylor Series Expansion

A Taylor series is a way to represent a function as an infinite sum of terms, calculated from the values of the function’s derivatives at a single point. For a function \(f(z)\) expanded around \(z = a\), the Taylor series is:
\[ f(z) = f(a) + f'(a)(z - a) + \frac{f''(a)}{2!}(z - a)^2 + \frac{f'''(a)}{3!}(z - a)^3 + ... \]
Each term involves higher powers of \(z - a\), scaled by the function’s derivatives at point \(a\).
In our specific problem, we expanded \(\text{cosh}(z)\) using its Taylor series at \(z = 0\). This provided a straightforward method to rewrite the numerator of our function as a sum of polynomial terms. Because mathematical series can be broken down and then reassembled to highlight different aspects of function behavior, this manipulation with the Taylor series made it easy to proceed with obtaining the Laurent series and ultimately, the residue.

Residues at Poles

Residues are a central concept in complex analysis, particularly useful in evaluating complex integrals via the Residue Theorem. Simply put, the residue of a function at a pole is the coefficient of the \(\frac{1}{z}\) term in its Laurent series expansion around that pole.
Poles are types of singularities where a function heads towards infinity as \(z\) approaches a specific point. If \(f(z)\) has a pole at \(z = c\), we can find its residue by focusing on how \(f(z)\) behaves in the Laurent series:
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