Problem 5 - Complex Analysis

Question

For the following functions \(f\) defined in a neighborhood of the point \(a \in \mathbb{C}\) determine the TAYLOR series at \(a\) and the convergence radius: (a) \(f(z)=\exp (z), \quad a=1\), (b) \(f(z)=\frac{1}{z}\), (c) \(f(z)=\frac{1}{z^{2}-5 z+6}, \quad a=0\) (d) \(f(z)=\frac{1}{(z-1)(z-2)}, \quad a=0\)

Step-by-Step Solution

Answer

(a) Series: \( e\sum_{n=0}^{\infty} \frac{(z-1)^n}{n!} \), \( R = \infty \). (b) No series, pole at 0. (c) \( R = 2 \). (d) \( R = 1 \).

Step 1: Understanding the Problem

We are given a set of functions and need to find their Taylor series around specified points along with determining the radius of convergence for each series.

Step 2: Function (a) - Taylor Series for \( f(z) = \exp(z) \) at \( a = 1 \)

The exponential function \( \exp(z) \) has a Taylor series expansion \( \sum_{n=0}^{\infty} \frac{z^n}{n!} \). To find the series at \( a = 1 \), use \( z = w + 1 \). Thus, \( \exp(w+1) = e \exp(w) \) and the Taylor series at \( a = 1 \) becomes \( e \sum_{n=0}^{\infty} \frac{(z-1)^n}{n!} \). The exponential function is entire, so the radius of convergence is infinite (\( R = \infty \)).

Step 3: Function (b) - Taylor Series for \( f(z) = \frac{1}{z} \) at \( a = 0 \)

The function \( f(z) = \frac{1}{z} \) is not defined at \( z = 0 \). For a Taylor series centered at zero, check the function's expansions in terms of \( z \). The Laurent series is used due to the singularity, given by \( \sum_{n=0}^{\infty} -z^{-n-1} \). In this case, there isn’t a regular Taylor series around zero due to a pole. The 'radius of convergence' formality does not apply here due to the pole at \( z = 0 \).

Step 4: Function (c) - Taylor Series for \( f(z) = \frac{1}{z^2 - 5z + 6} \) at \( a = 0 \)

First, factor the denominator: \( z^2 - 5z + 6 = (z-2)(z-3) \). Partial fractions give: \( \frac{1}{z^2 - 5z + 6} = \frac{A}{z-2} + \frac{B}{z-3} \). Solving, \( A = -1 \) and \( B = 1 \), hence \( \frac{-1}{z-2} + \frac{1}{z-3} \). For the series around 0, expand each fraction. The series converges as \( |z| < 2 \), since the singularity closest to 0 is \( z = 2 \). Thus, the radius of convergence \( R = 2 \).

Step 5: Function (d) - Taylor Series for \( f(z) = \frac{1}{(z-1)(z-2)} \) at \( a = 0 \)

Similarly, apply partial fraction decomposition: \( \frac{1}{(z-1)(z-2)} = \frac{1}{z-1} - \frac{1}{z-2} \). Expand each term in series format around \( z = 0 \). The convergence is valid within the circle \( |z| < 1 \) due to the nearest pole at \( z = 1 \). Therefore, the radius of convergence \( R = 1 \).

Key Concepts

Radius of Convergence

The radius of convergence is a fundamental concept when dealing with power series, such as those you encounter in Taylor expansions. It's denoted as \( R \) and represents the largest distance from the center of the series within which the series converges.
Imagine plotting each term of a power series as a circle centered on a point \( a \) (our expansion point). Every "circle" corresponds to an area where the series converges. The radius of this circle is our radius of convergence.

• For entire functions like the exponential function \( \exp(z) \), the series converges everywhere, giving an infinite radius of convergence, \( R = \infty \).
• For rational functions or functions with poles, the radius of convergence is determined by the nearest pole to the point \( a \).
• This is because a power series cannot converge at the poles of the function it represents.

Understanding how to calculate the radius of convergence can give insight into where a power series is applicable and is crucial when working with complex functions.

Complex Functions

Complex functions extend the concept of functions to the complex plane, allowing the use of complex numbers. Understanding these is crucial for series expansions, as many functions have singularities or peculiar behavior in the complex plane.
In the given exercises, several types of functions are considered:

• Polynomials and rational functions reveal their behavior through their zeros and poles.
• The exponential function, for example, is an entire function, meaning it's differentiable everywhere in the complex plane.
• Rational functions like \( \frac{1}{z} \) or \( \frac{1}{z-1} \) exhibit singular points (poles), where the function takes on infinite values.

These points dramatically affect the series expansions and indicate where series forms like Taylor or Laurent can be validly applied. Mastering this allows students to delve deeper into the behavior and properties of complex functions.

Partial Fraction Decomposition

Partial fraction decomposition is a valuable tool for breaking down complex rational expressions into simpler components. This technique is essential when dealing with functions that need to be expanded around a particular point, especially in complex analysis.
For example, consider the function \( f(z) = \frac{1}{z^2 - 5z + 6} \). To express it in a more "workable" form, you first factor the denominator: \( (z-2)(z-3) \). Through partial fraction decomposition, this then splits into simpler fractions, \( \frac{A}{z-2} + \frac{B}{z-3} \), which can each be expanded separately.

• This method simplifies complex expressions into forms where a power series expansion can be easily applied.
• Each term can be individually expanded in their respective regions of convergence.

Partial Fraction Decomposition thus turns a seemingly insurmountable function into manageable pieces, a crucial step in effectively mastering series solutions in complex function problems.