Problem 25 - Advanced Engineering Mathematics

Question

In Problems 25 and 26, use partial fractions as an aid in obtaining the Maclaurin series for the given function. Give the radius of convergence of the series. $$ f(z)=\frac{i}{(z-i)(z-2 i)} $$

Step-by-Step Solution

Answer

The Maclaurin series is obtained by expanding each term in partial fractions. Radius of convergence is 1.

Step 1: Express in Partial Fraction Form

The function given is \( f(z)=\frac{i}{(z-i)(z-2i)} \). Start by expressing this in partial fraction form: \( \frac{A}{z-i} + \frac{B}{z-2i} \). Solve for constants \( A \) and \( B \) by writing: \( i = A(z-2i) + B(z-i) \).

Step 2: Solve for Coefficients A and B

To find \( A \) and \( B \), equate coefficients. Set \( z = i \) to isolate \( B \): \( i = B(i-i) + A(i-2i) \). Since \( i - i = 0 \), this equation simplifies to \( i = A(-i) \). Thus, \( A = -i \). Set \( z = 2i \) to isolate \( A \): \( i = A(2i-i) + B(2i-i) \). This gives \( i = B(i) \), so \( B = 1 \).

Step 3: Formulate f(z) with A and B

Substitute back: \( f(z)=\frac{i}{(z-i)(z-2i)} = \frac{-i}{z-i} + \frac{1}{z-2i} \).

Step 4: Find Maclaurin Series for Each Term

Expand each term around \( z = 0 \). The Maclaurin series for \( \frac{1}{1-u} \) is \( 1 + u + u^2 + u^3 + \cdots \). For \( \frac{-i}{z-i} \), set \( u = \frac{z}{i} \) for \( \frac{-i}{z-i} = \frac{-1}{1 - \frac{z}{i}} \), which yields \( -1 - \frac{z}{i} - \frac{z^2}{i^2} - \cdots \). For \( \frac{1}{z-2i} \), set \( u = \frac{z}{2i} \), which gives \( \frac{-1}{1 - \frac{z}{2i}} = 1 + \frac{z}{2i} + \frac{z^2}{(2i)^2} + \cdots \).

Step 5: Combine Series

Combine the series obtained: \(-1 - \frac{z}{i} - \frac{z^2}{i^2} - \cdots + 1 + \frac{z}{2i} + \frac{z^2}{4i^2} + \cdots\). Simplify by combining like terms. Each coefficient can be adjusted by common factors of powers of \( z \).

Step 6: Determine Radius of Convergence

The radius of convergence \( R \) is determined by the smallest denominator in the partial fraction form which is min(1, 2) for \( z=i, z=2i \), thus \( R = 1 \).

Key Concepts

Partial Fractions

To understand the given exercise, we must first explore the concept of partial fractions. Partial fraction decomposition is a method used in algebra to break down complex rational expressions into simpler fractions. Imagine you have a complicated fraction, and we want to express it as a sum of simpler fractions. This process is especially useful in calculus, as simpler fractions are easier to integrate or differentiate.
In the problem, we have the function:

• \( f(z) = \frac{i}{(z-i)(z-2i)} \)

By expressing it in partial fraction form, we instead write:

• \( \frac{A}{z-i} + \frac{B}{z-2i} \)

Through algebra, we solve for the values of constants \( A \) and \( B \), making the function easier to work with. This form allows us to evaluate and manipulate the function intuitively. Partial fractions are a fundamental tool in many mathematical applications, including series expansions and solving differential equations.

Radius of Convergence

The radius of convergence pertains to the interval within which a given power series converges. For a complex function expressed in terms of its Maclaurin series, this determines how far from the center (usually zero) we can go without the series diverging.
In the current exercise, once split into partial fractions, the series expansion for each term gives insights into convergence. From the step by step solution, the radii are based on the distance of singularities from the expansion point. We have singularities at \( z = i \) and \( z = 2i \), meaning our radius of convergence \( R \) will be the smallest distance between zero and these singularities:

• \( R = \min(1, 2) = 1 \)

Therefore, the series will converge for all \( z \) where \( |z| < 1 \). Understanding the radius of convergence is crucial because it outlines the valid region for using the power series as an accurate representation of the function.

Complex Functions

In mathematics, complex functions are functions that have complex numbers as their inputs. These functions can exhibit behaviors quite different from real-valued functions, particularly because they can have discontinuities or poles (singularities) within the complex plane.
The function in the exercise is given as:

• \( f(z) = \frac{i}{(z-i)(z-2i)} \)

This is a complex function due to its complex variables and coefficients. When dealing with complex functions, tools like partial fractions and convergence radii become essential.

• Partial fractions help in simplifying the function into smaller parts that are easier to expand as a series.
• The radius of convergence identifies the domain where the series representation of the function is valid.

Mastering complex functions involves understanding both algebraic manipulations and the geometric interpretations, which can be visualized in the complex plane.