Problem 1 - A First Course in Complex Analysis with Applications

Question

In Problems 1-12, use known results to expand the given function in a Maclaurin series. Give the radius of convergence \(R\) of each series. $$ f(z)=\frac{z}{1+z} $$

Step-by-Step Solution

Answer

Maclaurin series: \( \sum_{n=0}^{\infty} (-1)^n z^{n+1} \); Radius of convergence \( R = 1 \).

Step 1: Understand the Maclaurin Series

A Maclaurin series is a type of Taylor series expansion where the series is centered at zero. For a function \( f(z) \), the Maclaurin series is given by \( f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \ldots \). We seek to express \( \frac{z}{1+z} \) in this form.

Step 2: Re-express the Function

The function \( \frac{z}{1+z} \) can be viewed as \( z \cdot \frac{1}{1+z} \). Recognize \( \frac{1}{1+z} \) as a geometric series \( \frac{1}{1-(-z)} = \sum_{n=0}^{\infty} (-1)^n z^n \) for \( |z| < 1 \).

Step 3: Multiply by z

Multiply the expanded geometric series by \( z \):\[ \frac{z}{1+z} = z \left( \sum_{n=0}^{\infty} (-1)^n z^n \right) = \sum_{n=0}^{\infty} (-1)^n z^{n+1} \]This gives the Maclaurin series for \( \frac{z}{1+z} \).

Step 4: Identify the Radius of Convergence

The original geometric series \( \frac{1}{1+z} \) converges for \( |z| < 1 \). Thus, the series for \( \frac{z}{1+z} \) also converges for \( |z| < 1 \), giving a radius of convergence \( R = 1 \).

Key Concepts

Taylor Series

A Taylor series is a fundamental tool in calculus used to approximate functions with an infinite sum of polynomial terms. It is expressed in terms of the derivatives of a function at a single point. The general form of the Taylor series for a function \( f(z) \) centered at \( a \) is:

• \( f(a) + f'(a)(z-a) + \frac{f''(a)}{2!}(z-a)^2 + \frac{f'''(a)}{3!}(z-a)^3 + \ldots \)

A Maclaurin series is a special case of the Taylor series centered at zero, making it simpler to work with:

• \( f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \ldots \)

When we expand the function \( \frac{z}{1+z} \) using the Maclaurin series, we seek to express this function as an infinite series of terms involving powers of \( z \). This approach is useful for approximating functions and solving complex calculus problems by converting them into manageable polynomial forms.

Geometric Series

The geometric series is one of the simplest and most elegant series in mathematics. It sums the terms of the form \( a + ar + ar^2 + ar^3 + \ldots \) where \( a \) is the first term and \( r \) is the common ratio.

• The series converges if the absolute value of the common ratio is less than one, \(|r| < 1\). The sum is given by \( \frac{a}{1-r} \).

In this exercise, \( \frac{1}{1+z} \) is re-expressed as a geometric series:

• \( \frac{1}{1+z} = \frac{1}{1-(-z)} = \sum_{n=0}^{\infty} (-1)^n z^n \)

This representation is valid for \( |z| < 1 \). By recognizing this, we transform \( f(z) = \frac{z}{1+z} \) into \( z \cdot \frac{1}{1+z} \), leading to the Maclaurin series \( \sum_{n=0}^{\infty} (-1)^n z^{n+1} \). Understanding geometric series allows us to rewrite complex fractions as neat polynomial series.

Radius of Convergence

The radius of convergence is a critical concept when dealing with power series, such as the Taylor or Maclaurin series. It defines the interval within which a series converges, meaning the sum of the series approaches a limit as the number of terms increases.

• For a series centered at zero, convergence depends on the values of \( z \) where the series terms shrink towards zero.

In our problem, the series \( \frac{z}{1+z} = z \sum_{n=0}^{\infty} (-1)^n z^n = \sum_{n=0}^{\infty} (-1)^n z^{n+1} \) has a radius of convergence derived directly from the conditions of the geometric series:

• The original geometric series \( \frac{1}{1+z} = \sum_{n=0}^{\infty} (-1)^n z^n \) converges for \( |z| < 1 \).

Thus, the transformed Maclaurin series for \( \frac{z}{1+z} \) inherits the same convergence condition. Therefore, the radius of convergence is \( R = 1 \). Understanding the radius of convergence allows us to determine where our power series representation accurately approximates the function and, hence, is reliable for calculations.