Problem 13 - Complex Analysis for Mathematics and Engineering

Question

Use the Maclaurin series for \(\sin z\) and then long division to get the Laurent series for \(\csc z\) with \(\alpha=0\).

Step-by-Step Solution

Answer

The Laurent series for \(\csc z\) is \( \frac{1}{z} + \frac{z}{6} + \frac{7z^3}{360} + \cdots \).

Step 1: Identify the Maclaurin Series for \(\sin z\)

The Maclaurin series for \(\sin z\) is well-known and given by:\[sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots\]This series represents \(\sin z\) as a power series centered at 0.

Step 2: Express \(\csc z\) in Terms of \(\sin z\)

Recall that \(\csc z = \frac{1}{\sin z}\). We want to find the Laurent series for this expression about \(z = 0\). This requires performing long division of the Maclaurin series of \(\sin z\) into 1.

Step 3: Perform Long Division

To find \(\csc z\), perform long division by dividing 1 by the Maclaurin series of \(\sin z\).- Start by considering "1 divided by z", which gives the first term of the Laurent series: \(\frac{1}{z}\).- The result of dividing \(1\) by \(z\) minus subsequent terms leads to:\[\csc z = \frac{1}{z} + \frac{z}{6} + \frac{7z^3}{360} + \cdots\]

Step 4: Combine and Simplify the Series

The division continues by multiplying the series \(z - \frac{z^3}{6} + \cdots\) into \(-\frac{1}{z}\) and simplifying each term. Subsequent terms follow similarly, leading ultimately to the Laurent series expansion of \(\csc z\).The final index includes both positive and negative powers. The result is:\[\csc z = \frac{1}{z} + \frac{z}{6} + \frac{7z^3}{360} + \cdots\]

Key Concepts

Maclaurin series

The Maclaurin series is a special case of the Taylor series expansion and is an essential tool in mathematics to express functions as infinite sums of their derivatives at a single point, usually at zero. This power series makes it easier to analyze and compute functions, especially in calculus and complex analysis.

When we talk about the Maclaurin series for \( \sin z \), it becomes essential to remember that it represents \( \sin z \) as a series centered at \( z = 0 \). The Maclaurin series for \( \sin z \) is:

• \( \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots \)

This series expands infinitely using odd powers of \( z \) and alternates between positive and negative terms, capturing the periodic nature of the sine function. Understanding and deriving Maclaurin series is highly useful for approximating functions around zero, where many real-world applications rely on such simplifications. Maclaurin series often pave the way to solving more complex problems, such as deriving the Laurent series in this exercise.

long division

Long division is a method used to divide one polynomial or series by another. In this exercise, it helps us find the Laurent series for \( \csc z \) using the Maclaurin series of \( \sin z \).

The Laurent series is a type of series similar to the Taylor series, but includes terms with negative powers. To find the Laurent series for \( \csc z \), we start by dividing 1 by the Maclaurin series of \( \sin z \).

Let's break down how long division is applied:

• Begin with "1 divided by \( z \)", the leading term of our series, giving us a starting point of \( \frac{1}{z} \).
• Subsequent division steps involve dividing the remainder by \( z \) and the following terms: \( - \frac{z^3}{3!} \), \( \frac{z^5}{5!} \), and so forth.

Through this iterative process, long division reveals the Laurent series:

• \( \csc z = \frac{1}{z} + \frac{z}{6} + \frac{7z^3}{360} + \cdots \)

Each division refines the series, providing more accuracy and insight into \( \csc z \) around \( z = 0 \). Long division of series is a fundamental technique in complex functions and helps solve otherwise difficult mathematical problems.

sin z

The function \( \sin z \) is a fundamental trigonometric function that describes the sine of an angle \( z \), commonly encountered in various fields of mathematics and physics. As a periodic function, \( \sin z \) exhibits a wave-like behavior that repeats every \( 2\pi \) radians.

The significance of \( \sin z \) in this exercise is to express it through its infinite Maclaurin series. This expression becomes a crucial stepping stone in deriving other functions, such as \( \csc z \), the reciprocal of \( \sin z \). By understanding the power series expansion:

• \( \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots \)

we lay the groundwork for performing further mathematical operations, such as series division, to obtain the Laurent series.

Knowing the behavior of \( \sin z \) breakdown into its series form enables mathematicians and students alike to manipulate these expressions for various applications, such as solving differential equations, optimizing engineering systems, or in the case of this exercise, finding the corresponding function \( \csc z \) effectively. Understanding \( \sin z \) and its properties can broaden your appreciation of trigonometry and its interconnectedness with other mathematical functions.