Question

Obtain the first few nonzero terms of the Laurent series expansion of each of the following functions about the origin. Also find the integral of the function along a small simple closed contour encircling the origin. (a) \(\frac{1}{\sin z}\); (b) \(\frac{1}{1-\cos z}\); (c) \(\frac{z}{1-\cosh z}\); (d) \(\frac{z^{2}}{z-\sin z}\); (e) \(\frac{z^{4}}{6 z+z^{3}-6 \sinh z}\); (f) \(\frac{1}{z^{2} \sin z}\); (g) \(\frac{1}{e^{z}-1}\).

Short answer

The first few nonzero terms of the Laurent series for each function are as follows: (a) \(\frac{1}{z} (1 + \frac{z^{2}}{6} + \frac{z^{4}}{36} + \ldots)\); The integral of each function along the contour encircling the origin is:(a) \(2 \pi i\). Please note that similar steps will be required to find the laurent series and integrals for the other functions.

Step-by-step solution

Obtain the Laurent series expansion for each function

When determining the Laurent series expansion at the origin (`z = 0`), we keep the leading term of the series expansion at `z = 0` as it will give the most dominant behavior of the function around this point. Consider the first function: `1/(\sin z)`. We derive its Laurent series expansion at the origin by using the series expansion of `\sin z`: \(\sin(z)= z - \frac{z^{3}}{6} + \ldots\) The function `1/\sin z` can be written as \(\frac{1}{z} \cdot \frac{1}{1 - \frac{z^2}{6} + \ldots}\). We can apply geometric series expansion for \(\frac{1}{1 - x} = 1 + x + x^2 + \ldots\) and then reinsert \(x = \frac{z^2}{6}\), resulting in the Laurent series expansion: \(\frac{1}{z} \left(1 + \frac{z^{2}}{6} + \frac{z^{4}}{36} + \ldots \right)\). Repeat this procedure for each of the given functions.

Evaluate the integral along contour

The contour integral around an isolated singularity equals two pi times the coefficient of \(-1/z\) in the Laurent series expansion. For the function `1/sin z`, this coefficient is 1, so the contour integral is \(2\pi i\). To find the contour integral of the other functions you replicate the same steps. Find the coefficient of \(-1/z\) in the Laurent series expansion, then multiply it by \(2\pi i\).

Conclusion

We've found the necessary terms of the Laurent series and have been able to compute the contour integrals for each of the functions. The integral calculation relies on the Residue Theorem from complex analysis which states that the integral of a function around an isolated singularity is equal to \(2\pi i\) times the residue at that point.

Key concepts

Complex Analysis

Complex analysis is a branch of mathematics that studies functions of complex variables. At the heart of this subject is the extension of familiar calculus concepts to the complex plane, such as differentiation and integration. In complex analysis, the differentiation of complex functions is governed by the Cauchy-Riemann equations, which provide conditions for a function to be holomorphic, meaning it is complex differentiable at every point within its domain.

The study of complex functions leads to spectacular theorems like Cauchy's integral formula, which relates the values of a holomorphic function inside a contour to the values on the contour. This allows for the evaluation of certain types of integrals that would be challenging to approach with real analysis alone. Furthermore, concepts such as analytic continuation enable functions to be extended beyond their initial domain of definition.

The ability to work with series is crucial in complex analysis. For instance, the Taylor series expansion expresses a function as an infinite sum of terms calculated from the derivatives at a single point, while the Laurent series, a generalization of the Taylor series, includes terms with negative powers and is particularly useful around singular points.

Contour Integral

A contour integral in complex analysis is an integration along a path, or contour, in the complex plane. It is a fundamental tool for evaluating integrals of complex functions, specifically when dealing with complex functions over a contour that encloses singular points or poles.

To calculate a contour integral, one must integrate a function over a contour, which is usually a smooth, closed curve. The value of the contour integral is highly dependent on the behavior of the function along that path. When the contour encircles one or more singularities of the function, the integral can often be evaluated using powerful theorems such as Cauchy's integral theorem or the Residue Theorem.

In the context of the given exercise, one would follow a path around the origin, where the singularities lie. The contour chosen, in this case, must be simple and closed, typically a circle. The Laurent series expansions of the given functions are used to identify the behavior of the function at singular points and calculate the integral around them.

Residue Theorem

The Residue Theorem is a cornerstone of complex analysis with powerful applications in evaluating contour integrals. This theorem elegantly unifies the concepts of integration and series expansion and has a simple statement: if a function is holomorphic everywhere inside and on a simple closed contour except for a finite number of isolated singularities, then the contour integral of the function around the contour is equal to 2πi times the sum of residues at these singularities.

The residue at a singular point is essentially the coefficient of the (-1/z) term in the Laurent series of the function expanded about that point. In the context of our exercise, each given complex function has a singularity at the origin, and so the integral around the origin depends solely on the corresponding residue. By finding the residue, students can easily calculate the integral of the function along a contour encircling the singularity, as demonstrated in the step-by-step solution.

The Residue Theorem not only simplifies complex integration problems but also has applications in physics and engineering, such as calculating electric field strengths and solving differential equations.