Problem 49 - A First Course in Complex Analysis with Applications

Question

(a) Consider the function \(\operatorname{Ln}(1+z)\). What is the radius of the largest circle centered at the origin within which \(f\) is analytic. (b) Expand \(f\) in a Maclaurin series. What is the radius of convergence of this series? (c) Use the result in part (b) to find a Maclaurin series for \(\operatorname{Ln}(1-z)\). (d) Find a Maclaurin series for \(\operatorname{Ln}\left(\frac{1+z}{1-z}\right)\).

Step-by-Step Solution

Answer

(a) Radius = 1. (b) Radius of convergence = 1. (c) Series: \(-\sum_{n=1}^{\infty} \frac{z^n}{n}\). (d) Series: \(2 \sum_{n=0}^{\infty} \frac{z^{2n+1}}{2n+1}\).

Step 1: Determine the Region of Analyticity for \\operatorname{Ln}(1+z)

The function \( \operatorname{Ln}(1+z) \) is analytic within a circle centered at the origin where \( |z| < 1 \). This is because the natural logarithm function with respect to \( z \) is only defined when \( 1 + z eq 0 \), which implies \( z eq -1 \). Therefore, the circle of analyticity includes all values \( z \) such that \( |z| < 1 \).

Step 2: Find the Maclaurin Series for \\operatorname{Ln}(1+z)

The Maclaurin series expansion for a function \( f(z) \) about \( z = 0 \) is given by \( f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \). For \( \operatorname{Ln}(1+z) \), we recognize it as the standard series: \[ \operatorname{Ln}(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} z^n}{n} \].

Step 3: Determine the Radius of Convergence for \\operatorname{Ln}(1+z) Series

The radius of convergence \( R \) for a power series \( \sum a_n z^n \) is given by \( R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}} \). Here, the coefficients are \( a_n = \frac{(-1)^{n+1}}{n} \). Therefore, we find that \( \limsup |a_n|^{1/n} = 1 \), so the radius of convergence \( R = 1 \).

Step 4: Find the Maclaurin Series Expansion for \\operatorname{Ln}(1-z)

To find the series for \( \operatorname{Ln}(1-z) \), substitute \( z \) with \( -z \) in the series for \( \operatorname{Ln}(1+z) \):\[ \operatorname{Ln}(1-z) = -\left(z + \frac{z^2}{2} + \frac{z^3}{3} + \cdots \right) = -\sum_{n=1}^{\infty} \frac{z^n}{n} \].

Step 5: Calculate the Maclaurin Series for \\operatorname{Ln}\left(\frac{1+z}{1-z}\right)

Using the results from previous steps, apply the formula:\[ \operatorname{Ln}\left(\frac{1+z}{1-z}\right) = \operatorname{Ln}(1+z) - \operatorname{Ln}(1-z) \].Use the series derived:\[ \operatorname{Ln}\left(\frac{1+z}{1-z}\right) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} z^n}{n} - \left(-\sum_{n=1}^{\infty} \frac{z^n}{n}\right) = 2 \sum_{n=0}^{\infty} \frac{z^{2n+1}}{2n+1} \].

Key Concepts

Maclaurin Series

A Maclaurin series is a type of Taylor series expansion. It focuses on expanding a function at zero. This technique is a great tool for approximating functions that are smooth and have derivatives of all orders at zero. With a Maclaurin series, you can express a function as an infinite sum.For example, the function \( \operatorname{Ln}(1+z) \) can be expanded as:

• First term: \( z \)
• Second term: \( -\frac{z^2}{2} \)
• Third term: \( \frac{z^3}{3} \)
• And so on...

Using the formula:\[ \operatorname{Ln}(1+z) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} z^n}{n} \]Understanding this series helps to see the function as a long polynomial. This series helps in analyzing the behavior of functions within their convergence range.

Radius of Convergence

The radius of convergence is crucial when dealing with power series. It tells us the region for which the series converges to the function. In simpler terms, it defines a circle around zero within which the series works well.For the function \( \operatorname{Ln}(1+z) \), the radius of convergence is found using the formula:\[ R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}}\]Here, the series coefficients \( a_n \) are \( \frac{(-1)^{n+1}}{n} \). The solution shows that the limit superior, \( \limsup |a_n|^{1/n} \), equals 1. Therefore, the radius of convergence \( R = 1 \). This means the series converges when \( |z| < 1 \). Understanding the radius of convergence helps clarify where the series representation of a function is valid.

Analytic Functions

An analytic function is one that is smooth and differentiated at every point in its domain. The concept of analyticity is closely related to power series, as these functions can be expressed using such series within their region of analyticity.For the function \( \operatorname{Ln}(1+z) \), it is essential to find where it remains analytic. A function is analytic in a region if it can be represented by a convergent power series in that region. In this case, \( \operatorname{Ln}(1+z) \) is analytic within the circle defined by \( |z| < 1 \).This means the function can be represented by its Maclaurin series exactly in this region. Knowing when and where a function is analytic is key when working with complex functions, allowing precise approximations and solutions.