Problem 461 - Advanced Calculus

Question

Find the partial fraction expansion of $$ f(z)=\left\\{\left(z^{2}+1\right) /\left(z^{3}+4 z^{3}+3 z\right)\right\\} $$ using the theory of residues.

Step-by-Step Solution

Answer

The partial fraction expansion of \(f(z) = \frac{z^{2}+1}{z^{3}+4z^{2}+3z}\) using the theory of residues is: \[f(z) = \frac{1}{3z} - \frac{1}{z+1} - \frac{4}{3(z+3)}\]

Step 1: Factoring the denominator

Factor the denominator: \[D(z)=z^{3}+4z^{2}+3z = z(z^2 + 4z + 3) = z(z+1)(z+3)\]

Step 2: Finding the simple poles

The simple poles of the function are the roots of the denominator polynomial \(D(z)\). The denominator has roots at: \[z = 0, -1, -3\] So there are three simple poles: \(z=0, z=-1\), and \(z=-3\).

Step 3: Computing residues at each pole

Now we need to calculate the residues at each pole. Using the residue theorem for a simple pole, the residue for each pole is given by: \[Res(f(z), z = z_i) = \lim_{z \to z_i} (z - z_i) f(z)\] At the pole \(z = 0\): \[ \begin{aligned} Res(f(z), z = 0) &= \lim_{z \to 0} (z-0) \frac{z^2+1}{z(z+1)(z+3)} \\ &= \lim_{z \to 0} \frac{z^2+1}{(z+1)(z+3)} \\ &= \frac{1}{(0+1)(0+3)} = \frac{1}{3} \end{aligned} \] At the pole \(z = -1\): \[ \begin{aligned} Res(f(z), z = -1) &= \lim_{z \to -1} (z+1) \frac{z^2+1}{z(z+1)(z+3)} \\ &= \lim_{z \to -1} \frac{z^2+1}{z(z+3)} \\ &= \frac{(-1)^2 + 1}{(-1)(-1 + 3)} = -\frac{2}{2} = -1 \end{aligned} \] At the pole \(z = -3\): \[ \begin{aligned} Res(f(z), z = -3) &= \lim_{z \to -3} (z+3) \frac{z^2+1}{z(z+1)(z+3)} \\ &= \lim_{z \to -3} \frac{z^2+1}{z(z+1)} \\ &= -\frac{(-3)^2 + 1}{(-3)(-3 + 1)} = -\frac{8}{6} =-\frac{4}{3} \end{aligned} \]

Step 4: Constructing the partial fraction expansion

The partial fraction expansion of \(f(z)\) can be written as a sum of fractions, using the residues we calculated: \[f(z) = \frac{1}{3} \cdot \frac{1}{z} -1 \cdot \frac{1}{z+1} - \frac{4}{3} \cdot \frac{1}{z+3}\] Thus, the partial fraction expansion of \(f(z)\) is: \[f(z) = \frac{1}{3z} - \frac{1}{z+1} - \frac{4}{3(z+3)}\]

Key Concepts

Residue Theorem

The residue theorem is a powerful tool in complex analysis, particularly useful when dealing with integrals and series involving complex functions. It essentially states that if you have a complex function which is analytic in a region except for a few isolated singularities, the integral of the function over a closed contour within this region is directly related to the sum of residues at those singularities.

What exactly is a residue? In simple terms, a residue at a particular point is a coefficient that captures the behavior of a function near that singularity. For simple poles, which are singularities where the function goes to infinity in some direction but is finite in others, the residue can be calculated by multiplying the function by \(z - z_i\) (where \(z_i\) is the location of the pole) and then taking the limit as \(z\) approaches \(z_i\). This method was used in the given exercise to find the coefficients that would constitute the partial fraction expansion, enabling us to rewrite the complex function \(f(z)\) in a much simpler form.

Complex Poles

Complex poles are certain types of singularities in a complex function where the function's value becomes unbounded. In our context, when analyzing the function \(f(z)\) for partial fraction expansion, we seek poles that are zeros of the denominator. It's critical to identify these poles accurately, as they are pivotal in calculating the residues.

In the given solution, we discovered that \(z = 0, -1, -3\) are the simple poles of our function. Simple means that the function does not go to infinity faster than \(\frac{1}{z}\) as \(z\) approaches the pole. Knowing the locations of these poles allows us to proceed with calculating the residues, which are essential ingredients in the partial fraction decomposition of \(f(z)\). Each pole contributes a term to this expansion, reflecting the function's behavior near each singularity.

Factoring Polynomials

Factoring polynomials is a fundamental skill in both algebra and complex analysis. It's the process of breaking down a polynomial into a product of its factors, which are simpler polynomials that multiply to give the original one. In the realm of complex analysis, factoring polynomials is particularly valuable because it reveals the zeros of the polynomial, which correspond to poles of a rational function.

The problem we are considering presents us with a polynomial in the form of the denominator of \(f(z)\). Factoring this polynomial, as illustrated in Step 1 of the solution, is the first step towards identifying singularities and, consequently, for finding the residues. It simplifies our analysis and allows us to handle each singularity separately when applying the residue theorem. It's just like breaking a complex problem into smaller, manageable pieces.

Complex Analysis

Complex analysis is the study of complex numbers and functions involving them. It's an area of mathematics filled with beautiful theorems and practical applications. Functions of a complex variable are analogous to real functions, but they have unique properties due to the multidimensional nature of complex numbers.

One of the striking results of complex analysis is that complex differentiable functions are infinitely differentiable and have convergent power series expansions around points where they are analytic. The concepts of residues and poles come from this area of mathematics and are essential for understanding how functions behave near points where they aren't defined. The partial fraction expansion technique, pivotal in solving the given exercise, is driven by the principles of complex analysis. It enables us to express a complex rational function as a sum of simpler terms, making it easier to integrate, differentiate, or evaluate the function.