Problem 26 - A First Course in Complex Analysis with Applications

Question

Use partial fractions as an aid in obtaining the Maclaurin series for the given function. Give the radius of convergence \(R\) of the series. $$ f(z)=\frac{z-7}{z^{2}-2 z-3} $$

Step-by-Step Solution

Answer

Maclaurin series: \( \sum_{n=0}^{\infty} \left( \frac{-2}{3^{n+1}} + (-1)^{n+1} \right) z^n \). Radius of convergence \( R = 1 \).

Step 1: Partial Fraction Decomposition

First, decompose the given function \( f(z) = \frac{z-7}{z^2 - 2z - 3} \). Start by factoring the denominator: \( z^2 - 2z - 3 = (z-3)(z+1) \). Therefore, we can express \( f(z) \) as follows: \( \frac{z-7}{(z-3)(z+1)} = \frac{A}{z-3} + \frac{B}{z+1} \). To find \( A \) and \( B \), use the cover-up method or equate coefficients after multiplying through by the denominator.

Step 2: Solve for Coefficients A and B

Equate \( z-7 = A(z+1) + B(z-3) \). Expand and collect like terms: \( A(z+1) + B(z-3) = Az + A + Bz - 3B = (A+B)z + (A-3B) \). Equate coefficients with \( z-7 \):\(A+B = 1\) and \(A-3B=-7\). Solve these equations to find \(A = 2\) and \(B = -1\).

Step 3: Express the Function with Coefficients

Now we have: \( \frac{z-7}{(z-3)(z+1)} = \frac{2}{z-3} - \frac{1}{z+1} \). Next, express each term as a power series valid at \(z=0\).

Step 4: Find Maclaurin Series for Each Term

For \( \frac{2}{z-3} = \frac{-2}{3-z}\), use the series \( \frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \) with \( u = \frac{z}{3} \):\( \frac{-2}{3} \sum_{n=0}^{\infty} \left(\frac{z}{3}\right)^n = \sum_{n=0}^{\infty} \frac{-2}{3^{n+1}} z^n \). For \( \frac{-1}{z+1} \), use \( \frac{1}{1+v} = \sum_{n=0}^{\infty} (-1)^n v^n \) with \( v = -z \): \( \sum_{n=0}^{\infty} (-1)^{n+1} z^n \).

Step 5: Combine Series and Find Result

Combine the results: \( \sum_{n=0}^{\infty} \left( \frac{-2}{3^{n+1}} + (-1)^{n+1} \right) z^n \). This is the Maclaurin series representation of \( f(z) \).

Step 6: Determine the Radius of Convergence

The series \( \frac{-2}{3-z} \) converges for \( |z/3|<1 \), i.e., \(|z|<3\). The series \( \frac{-1}{z+1} \) converges for \(|z|<1\). The overall radius of convergence is where both series converge, thus \( R = 1 \).

Step 7: Conclusion

The Maclaurin series for the function is \( \sum_{n=0}^{\infty} \left( \frac{-2}{3^{n+1}} + (-1)^{n+1} \right) z^n \) with the radius of convergence \( R = 1 \).

Key Concepts

Maclaurin Series

The Maclaurin Series is a powerful tool used in complex analysis to express functions as infinite series. This expands a function into terms of powers of the variable around zero. If you have a function, like our example function, you can break it down into simpler components that are easier to work with. A Maclaurin Series takes the form:

• \( f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \)

Here, each term consists of the derivatives of the function evaluated at zero. This is a special case of the Taylor Series, expanded at point zero. As you begin to understand more about these series, you'll find they help in approximating complex functions, making calculations more manageable.
In this exercise, we used the decomposition into simpler fractions to separately compute series for each fraction. This allowed us to express the resulting function in a form that is an infinite sum, making it both easier to handle and to analyze mathematically.

Partial Fraction Decomposition

Partial Fraction Decomposition is a method used in simplifying complex rational expressions. When dealing with functions like \( \frac{z-7}{z^2 - 2z - 3} \), breaking it down into simpler fractions makes it easier to use in further calculations.

• You start by factoring the denominator.
• This allows you to express the function as a sum of fractions with simpler denominators.

We see that the denominator \(z^2 - 2z - 3\) can be factored into \((z-3)(z+1)\). From this, decompose the function using unknown constants \(A\) and \(B\) such that:

• \( \frac{z-7}{(z-3)(z+1)} = \frac{A}{z-3} + \frac{B}{z+1} \)

Finding these constants involves solving a system of equations, which results from equating the original function with its partial fraction form. This step provides a simpler representation for computation of the series.

Radius of Convergence

The Radius of Convergence is a critical concept when working with series, indicating where a series representation of a function converges. It's essential when examining the validity of a series solution, showing the values for which the series holds true.

• For each component series in the partial fractions, determine its radius.
• The smallest radius ensures overall convergence.

In our example:- The series from \( \frac{2}{z-3} \) converges when \(|z| < 3\).- The series from \( \frac{-1}{z+1} \) converges when \(|z| < 1\).The overall radius of convergence for the combined series is 1, as it is the most restrictive value among the components. Understanding this enables us to determine the application range for the Maclaurin Series, ensuring accurate function approximation within this domain.