Problem 30 - A First Course in Complex Analysis with Applications

Question

Use \(\cos z=1-\frac{z^{2}}{2 !}+\frac{z^{4}}{4 !}-\cdots, \sin z=z-\frac{z^{3}}{3 !}+\frac{z^{5}}{5 !}-\cdots\), and long division to find the first three nonzero terms of a Laurent series of the given function \(f\) valid for \(0<|z|<\pi\) $$ f(z)=\cot z $$

Step-by-Step Solution

Answer

The first three nonzero terms of the Laurent series for \(\cot z\) are \(\frac{1}{z} + \frac{z}{6} + \frac{z^3}{36}\).

Step 1: Express Cotangent

First, express \(f(z) = \cot z = \frac{\cos z}{\sin z}\) as a ratio of the cosine series to the sine series.

Step 2: Series for Cosine and Sine

Use the given series for cosine and sine: \[ \cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots \] \[ \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots \].

Step 3: Perform Long Division

Divide the series expansion of \(\cos z\) by the series of \(\sin z\), similar to polynomial long division. 1. Divide the first term of the numerator by the first term of the denominator: \( \frac{1}{z} = \frac{1}{z} \).2. Multiply \( \frac{1}{z} \) by \(\sin z\): \( 1 - \frac{z^2}{3!} + \cdots \).3. Subtract this result from \(\cos z\):\[ \left( 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots \right) - \left( 1 - \frac{z^2}{3!} + \cdots \right) = \frac{z^2}{6} - \frac{z^4}{24} + \cdots \].4. Divide \(\frac{z^2}{6}\) by \(z\): \(\frac{z}{6}\).5. Multiply \( \frac{z}{6} \) by \(\sin z\): \(\frac{z}{6} - \frac{z^3}{36} + \cdots \).6. Subtract from the remaining series: \[ \left( \frac{z^2}{6} - \frac{z^4}{24} \right) - \left( \frac{z^2}{6} - \frac{z^3}{36} \right) \].7. Result is \( \frac{z^3}{36} - \cdots \) implying next term in the quotient is \(\frac{1}{36}\).

Step 4: Gather the Terms

Collect the terms of the quotients from the division: \\[\cot z \approx \frac{1}{z} + \frac{z}{6} + \frac{z^3}{36} + \cdots\].This gives the first three non-zero terms of the Laurent series.

Key Concepts

Complex Analysis

Complex analysis is a fascinating field of mathematics that deals with functions that have complex numbers as their inputs. This area of study is crucial because many real-world phenomena can be modeled more effectively using complex variables. The two most common components in complex analysis are the real part and the imaginary part of a complex number, written as \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit.

• Functions in complex analysis often need to be both continuous and differentiable, a property known as being "holomorphic."
• Important theorems, such as Cauchy's integral theorem and the residue theorem, are foundational in complex analysis.

For our exercise, we're particularly interested in expanding the cotangent function in terms of its Laurent series. A Laurent series is a representation that allows us to express functions using both positive and negative powers of a variable. It’s particularly useful near singularities, points where a standard Taylor series would fail.

Taylor Series

In mathematics, a Taylor series is a way to represent a function as an infinite sum of terms. These terms are calculated using the derivatives of the function at a single point, often centered around zero. Taylor series are instrumental in simplifying complex functions, making them easier to work with.

• Each term in a Taylor series involves a derivative of the function, a factorial, and a power of the variable.
• For example, the Taylor series of \( \ cos(z) \ \) and \( \ sin(z) \ \) are given by \[\cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots\] and \[\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots\].

These series are particularly useful in representing periodic functions in complex analysis. They allow us to handle trigonometric functions more efficiently when analyzing singularities and performing operations such as division, as seen in our exercise.

Cotangent Function

The cotangent function, represented as \( \cot z = \frac{\cos z}{\sin z} \), is the reciprocal of the tangent function. In complex analysis, expanding \( \cot z \) into a Laurent series allows us to better understand its behavior near singularities, particularly where \( \sin z \) equals zero.

• To expand it into a Laurent series, we start with the Taylor series for \( \cos z \) and \( \sin z \).
• Using polynomial long division, we divide the \( \cos z \) series by the \( \sin z \) series, obtaining terms with both positive and negative powers.
• The result is a series that captures the essence of \( \cot z \) over an interval where usual series expansions fail due to division by zero issues at the singular points.

Understanding these expansions is a powerful tool in complex analysis, providing deeper insights into the periodicity and symmetry of trigonometric functions in the complex plane.