2.3 Polynomials and Rational Functions

\begin{aligned} P (z) & = a_{0} + a_{1} z + a_{2} z^{2} + \dots + a_{n} z^{n} \\ P^{'} (z) & = a_{1} + 2 \cdot a_{2} z + \dots + n \cdot a_{n} z^{n - 1} \end{aligned}

Fundamental Theorem of Algebra

Every polynomial of degree $n$ has $n$ roots (counting the multiplicity)
Polynomial division:
$P (z) = D (z) Q (z) + R (z)$

If $α_{1}$ is root ( $P (α_{1})) = 0$

$P (z) = (z - α_{1}) Q (z) + R$ , $R$ must be degree 0 since we are dividing by ( $z - α_{1}$ )

$P (α_{1}) = (α_{1} - α_{1}) Q (α) + R = 0 \times Q (α) + R = R = 0$
$⟹ P (z) = (z - α_{1}) P_{1} (z)$

We can repeat this process until we get

P (z) = a_{n} (z - α_{1}) (z - α_{2}) \dots (z - α_{n})

Another method is:

\begin{aligned} P (z) - \underset{0}{\underset{⏟}{P (α_{1})}} & = a_{n} z^{n} + \dots + a_{1} z + a_{0} - (a_{n} α_{1}^{n} + \dots a_{1} α_{1} + a_{0}) \\ = a_{n} (z^{n} - α_{1}^{n}) + \dots + a_{1} (z - α_{1}) \end{aligned}

For any $k \geq 1$ ,

z^{k} - α_{1}^{k} = (z - α_{1}) (z^{k - 1} + z^{k - 2} α_{1} + \dots + α_{1}^{k - 1}) .

Note that:

\begin{aligned} \log P (z) & = a_{n} + \log (z - α_{1}) + \dots \log (z - α_{_{n}}) \\ ⟹ \frac{P^{'} (z)}{P (z)} = \frac{1}{z - α_{1}} + \dots + \frac{1}{z - α_{n}} \end{aligned}

Convex Hull

Convex Set:

A set is convex if the the line that joins any two points in a set is also inside the set.

Convex Hull:

Is the smallest convex set that completely encloses a given set of points
image-7.png|420x169

Convex Polygon:

When we have finite points a complex polygon is equivalent to the convex hull if its interior angles are less than $π$

A point is in a complex $n -$ polygon if we can write that point as

m_{1} a_{1} + m_{2} a_{2} + \dots + m_{n} a_{n}

where $a_{1}, \dots, a_{n}$ are the vertices and $\sum_{i = 1}^{n} m_{i} = 1$ , $m_{i} \geq 0$

Gauss-Lucas Theorem:

Let $P (z)$ be a complex polynomial with roots $α_{1}, α_{2}, \dots, α_{n} \in C$ . Then the roots of $P^{'} (z)$ lie in the convex hull of $α_{1}, α_{2}, \dots, α_{n}$

Proof:

We know that $\frac{P^{'} (z)}{P (z)} = \sum_{i = 1}^{n} \frac{1}{z - α_{i}}$

Assume that $P^{'} (z_{0}) = 0 ⟹$

\begin{aligned} 0 = \sum_{i = 1}^{n} \frac{1}{z_{0} - α_{i}} \\ ⟹ 0 = \sum_{i = 1}^{n} \frac{\overset{―}{z_{0} - α_{i}}}{| z_{0} - α_{i} |^{2}} \\ ⟹ \sum_{i = 1}^{n} \frac{\overset{―}{z_{0}}}{| z_{0} - α_{i} |^{2}} = \sum_{i = 1}^{n} \frac{\overset{―}{α_{i}}}{| z_{0} - α_{i} |^{2}} \\ ⟹ \overset{―}{z_{0}} \cdot \sum_{i = 1}^{n} \frac{1}{| z_{0} - α_{i} |^{2}} = \sum_{i = 1}^{n} \frac{\overset{―}{α_{i}}}{| z_{0} - α_{i} |^{2}} \\ ⟹ z_{0} = \frac{1}{\sum_{i = 1}^{n} \frac{1}{| z_{0} - α_{i} |^{2}}} \sum_{i = 1}^{n} \frac{α_{i}}{| z_{0} - α_{i} |^{2}} \end{aligned}

let $m_{i} = \frac{1}{\sum_{i = 1}^{n} \frac{1}{| z_{0} - α_{i} |^{2}}} \frac{1}{| z_{0} - α_{i} |^{2}}$

⟹ \sum_{}^{} m_{i} = {(\sum_{i = 1}^{n} \frac{1}{| z_{0} - α_{i} |^{2}})}^{- 1} (\sum_{i = 1}^{n} \frac{1}{| z_{0} - α_{_{i}} |^{2}}) = 1

Rational Functions

A rational function is a quotient of two polynomials

R (z) = \frac{P (z)}{Q (z)}

Assuming $P$ and $Q$ have no common factors.
The zeros of $Q (z)$ are the poles of $R (z)$ .

We can write:

$R (z) = G (z) + \frac{P_{1} (z)}{Q (z)}$ using long division where degree of $G$ = degree( $P$ )-degree( $Q$ )

$P (z) = a_{n} z^{n} + a_{n - 1} z^{n - 1} + \dots + a_{0}$ ., substitute $z = 1 / z$ :

\begin{aligned} P (1 / z) & = a_{n} (1 / z)^{n} + a_{n - 1} (1 / z)^{n - 1} + \dots + a_{1} (1 / z) + a_{0} \\ P (1 / z) & = \frac{a_{n}}{z^{n}} + \frac{a_{n - 1}}{z^{n - 1}} + \dots + \frac{a_{1}}{z} + a_{0} \end{aligned}

To clear the fractions, let's find a common denominator is $z^{n}$ :

P (1 / z) = \frac{a_{n} + a_{n - 1} z + a_{n - 2} z^{2} + \dots + a_{1} z^{n - 1} + a_{0} z^{n}}{z^{n}} \overset{c a l l}{=} P_{1} (z)

Step 2: Analyze the Denominator $Q (1 / z)$

$Q (z) = b_{m} z^{m} + \dots + b_{0}$ .

Substitute $z = 1 / z$ and find the common denominator $z^{m}$ :

Q (1 / z) = \frac{b_{m} + b_{m - 1} z + \dots + b_{1} z^{m - 1} + b_{0} z^{m}}{z^{m}} \overset{c a l l}{=} Q_{1} (z)

Step 3: Combine them to find $R (1 / z)$

Now we just divide the two results:

R (1 / z) = \frac{P (1 / z)}{Q (1 / z)} = \frac{\frac{a_{n} + a_{n - 1} z + \dots + a_{0} z^{n}}{z^{n}}}{\frac{b_{m} + b_{m - 1} z + \dots + b_{0} z^{m}}{z^{m}}}

To simplify this complex fraction,

R (1 / z) = (\frac{a_{n} + a_{n - 1} z + \dots + a_{0} z^{n}}{z^{n}}) \cdot (\frac{z^{m}}{b_{m} + b_{m - 1} z + \dots + b_{0} z^{m}})

Now, just group the $z$ factors:

R (1 / z) = \frac{z^{m}}{z^{n}} \cdot \frac{a_{n} + a_{n - 1} z + \dots + a_{0} z^{n}}{b_{m} + b_{m - 1} z + \dots + b_{0} z^{m}}

R (1 / z) = z^{m - n} \cdot \frac{a_{n} + a_{n - 1} z + \dots + a_{0} z^{n}}{b_{m} + b_{m - 1} z + \dots + b_{0} z^{m}}

{\begin{cases} m > n, \infty is a 0 of order m - n \\ m < n, \infty is a pole of order n - m \\ m = n, R (\infty) = \frac{a_{n}}{b_{m}} \neq 0, \infty \end{cases}

Pole

A pole of $R (z)$ is a point where $R (z) = \infty$ or $Q (z) = 0$ but $P (z) \neq 0$

Order of a Pole

The order of the corresponding zero of $Q (z)$

If $α$ is a zero of $Q$ so that $Q (z) = (z - α)^{n} Q_{1} (z)$ and $α$ is not a root of $Q_{1} (z)$ then $n$ is the order of the pole.

The Partial Fraction Decomposition Theorem

Theorem: Any rational function $R (z) = P (z) / Q (z)$ can be uniquely decomposed into the form:

R (z) = G (z) + \sum_{j = 1}^{n} G_{j} (\frac{1}{z - z_{j}})

where $z_{1}, \dots, z_{n}$ are the distinct poles of $R (z)$ , $G (z)$ is a polynomial, and each $G_{j}$ is a polynomial without a constant term.

Proof

Step 1. Isolate the Polynomial Part (Behaviour at $\infty$ )

Using polynomial long division, we can always divide $P (z)$ by $Q (z)$ to get a quotient $G (z)$ and a remainder $P_{1} (z)$ :

R (z) = \frac{P (z)}{Q (z)} = G (z) + \frac{P_{1} (z)}{Q (z)}

By the definition of long division, $\deg (P_{1}) < \deg (Q)$ .

$G (z)$ is the polynomial part of $R (z)$ .
Let $R_{1} (z) = P_{1} (z) / Q (z)$ . This is a proper rational function, which means $lim_{z \to \infty} R_{1} (z) = 0$ .

Our new goal is to decompose the proper fraction $R_{1} (z)$ .

Step 2. Isolate the Behaviour at a Single Pole $z_{1}$

Let $z_{1}$ be a pole of $R_{1} (z)$ . We want to find the "principal part" $G_{1}$ associated with this pole.

Substitution: Perform a change of variables: $z = z_{1} + 1 / ζ$ .
This maps the pole $z \to z_{1}$ to the point $ζ \to \infty$ .
New Function: Define a new rational function $F (ζ)$ :

F (ζ) = R_{1} (z) = R_{1} (z_{1} + 1 / ζ)

Since $R_{1} (z)$ has a pole at $z_{1}$ , $F (ζ)$ must have a pole at $ζ = \infty$ .

Long Division (Again): A rational function with a pole at $\infty$ can be split by long division into its polynomial part and a proper remainder.

F (ζ) = G_{1} (ζ) + F_{1} (ζ)

where $G_{1} (ζ)$ is a polynomial and $F_{1} (ζ)$ is a proper rational function (so $lim_{ζ \to \infty} F_{1} (ζ) = 0$ ).

Check Constant Term: Since $R_{1} (z)$ is proper, $lim_{z \to \infty} R_{1} (z) = 0$ . In $ζ$ -coordinates, $z \to \infty$ corresponds to $ζ \to 0$ .

lim_{ζ \to 0} F (ζ) = lim_{ζ \to 0} (G_{1} (ζ) + F_{1} (ζ)) = lim_{z \to \infty} R_{1} (z) = 0

This implies $G_{1} (0) + F_{1} (0) = 0$ . This forces $G_{1} (ζ)$ to have no constant term (or $F_{1} (0)$ to cancel it). For simplicity, we define $G_{1}$ as the polynomial in $ζ$ without a constant term that captures the pole at $\infty$ .

Substitute Back: Now, replace $ζ$ with $1 / (z - z_{1})$ :

R_{1} (z) = F (\frac{1}{z - z_{1}}) = G_{1} (\frac{1}{z - z_{1}}) + F_{1} (\frac{1}{z - z_{1}})

Step 3. Analyze the New Remainder (Induction)

Let $H_{1} (z) = F_{1} (1 / (z - z_{1}))$ .

$G_{1} (1 / (z - z_{1}))$ is the principal part for the pole $z_{1}$ . It's a polynomial in $1 / (z - z_{1})$ , e.g., $\frac{c_{k}}{(z - z_{1})^{k}} + \dots + \frac{c_{1}}{z - z_{1}}$ .
What is $H_{1} (z)$ at the pole $z_{1}$ ? $$\lim_{z \to z_1} H_1(z) = \lim_{z \to z_1} F_1\left(\frac{1}{z-z_1}\right) = \lim_{\zeta \to \infty} F_1(\zeta) = 0$$

Since the limit of $H_{1} (z)$ at $z_{1}$ is finite (it's 0), $H_{1} (z)$ does not have a pole at $z_{1}$ .

We have successfully "peeled off" the pole at $z_{1}$ . We can now write:

R_{1} (z) = G_{1} (\frac{1}{z - z_{1}}) + H_{1} (z)

where $H_{1} (z)$ is a new rational function that contains all the other poles ( $z_{2}, \dots, z_{n}$ ), but is analytic at $z_{1}$ .

We repeat this process (inductively) for $H_{1} (z)$ and the pole $z_{2}$ , then for $H_{2} (z)$ and $z_{3}$ , and so on.

Step 4. The Final Remainder

After $n$ steps, we have "peeled off" all $n$ poles:

R_{1} (z) = G_{1} (\frac{1}{z - z_{1}}) + \dots + G_{n} (\frac{1}{z - z_{n}}) + H_{n} (z)

R_{1} (z) = \sum_{j = 1}^{n} G_{j} (\frac{1}{z - z_{j}}) + H_{n} (z)

This final remainder $H_{n} (z)$ is a rational function.

It has no finite poles (we removed them all). A rational function with no poles must be a polynomial.
It must be a proper fraction, since it's the sum/difference of other proper fractions.
- $lim_{z \to \infty} R_{1} (z) = 0$ (by definition).
- $lim_{z \to \infty} G_{j} (\frac{1}{z - z_{j}}) = G_{j} (0) = 0$ (since $G_{j}$ has no constant term).
- Therefore, $lim_{z \to \infty} H_{n} (z) = lim_{z \to \infty} (R_{1} (z) - \sum G_{j} (\frac{1}{z - z_{j}})) = 0 - 0 = 0$ .

A function $H_{n} (z)$ that is both a polynomial and tends to $0$ as $z \to \infty$ must be the zero polynomial: $H_{n} (z) = 0$ .

Conclusion

This leaves us with:

R_{1} (z) = \sum_{j = 1}^{n} G_{j} (\frac{1}{z - z_{j}})

Substituting this back into the result from Step 1 gives the final theorem:

R (z) = G (z) + R_{1} (z) = G (z) + \sum_{j = 1}^{n} G_{j} (\frac{1}{z - z_{j}})

Example

Decompose $R (z) = \frac{z^{4}}{(z - 1)^{3} (z + 1)}$

This fraction is improper because the degree of the numerator (4) is equal to the degree of the denominator ( $3 + 1 = 4$ ).

Target Form:

The decomposition will look like this:

R (z) = G (z) + \frac{A}{(z - 1)^{3}} + \frac{B}{(z - 1)^{2}} + \frac{C}{z - 1} + \frac{D}{z + 1}

$G (z)$ is the polynomial part.
$A, B, C$ are the coefficients for the pole at $z = 1$ (order 3).
$D$ is the coefficient for the pole at $z = - 1$ (order 1).

1. Polynomial Part $G (z)$

We use long division. First, expand the denominator:

$Q (z) = (z^{3} - 3 z^{2} + 3 z - 1) (z + 1) = z^{4} - 2 z^{3} + 2 z - 1$

Now, divide $z^{4}$ by this:

\begin{array}{r} 1 \\ z^{4} - 2 z^{3} + 0 z^{2} + 2 z - 1 \overset{―}{) z^{4} + 0 z^{3} + 0 z^{2} + 0 z + 0} \\ - (z^{4} - 2 z^{3} + 0 z^{2} + 2 z - 1) \\ \overset{―}{2 z^{3} + 0 z^{2} - 2 z + 1} \end{array}

$G (z) = 1$
The remainder is $R_{1} (z) = \frac{2 z^{3} - 2 z + 1}{(z - 1)^{3} (z + 1)}$

We now only need to decompose the remainder $R_{1} (z)$ .

2. Find Coefficient $D$ (Simple Pole $z = - 1$ )

We can use the simple "cover-up" method for the simple pole.

D = lim_{z \to - 1} (z + 1) R_{1} (z) = lim_{z \to - 1} \frac{2 z^{3} - 2 z + 1}{(z - 1)^{3}}

D = \frac{2 (- 1)^{3} - 2 (- 1) + 1}{(- 1 - 1)^{3}} = \frac{- 2 + 2 + 1}{(- 2)^{3}} = \frac{1}{- 8}

$D = - 1 / 8$

3. Find Coefficients $A, B, C$ (Order 3 Pole $z = 1$ )

We use the general "Taylor series" method.

Define $Φ (z)$ : "Clear" the pole by multiplying $R_{1} (z)$ by $(z - 1)^{3}$ :

Φ (z) = (z - 1)^{3} R_{1} (z) = \frac{2 z^{3} - 2 z + 1}{z + 1}

Match Forms: We know that...

Φ (z) = A + B (z - 1) + C (z - 1)^{2} + \dots

..and we also know from Taylor's theorem that...

Φ (z) = Φ (1) + \frac{Φ^{'} (1)}{1!} (z - 1) + \frac{Φ^{″} (1)}{2!} (z - 1)^{2} + \dots

Find Coefficients:
- $A = Φ (1)$
- $B = Φ^{'} (1) / 1!$
- $C = Φ^{″} (1) / 2!$

Let's calculate these:

To find $A$ :

A = Φ (1) = \frac{2 (1)^{3} - 2 (1) + 1}{1 + 1} = \frac{1}{2}

$A = 1 / 2$

To find $B$ :

First, find $Φ^{'} (z)$ using the quotient rule on $\frac{2 z^{3} - 2 z + 1}{z + 1}$ :

Φ^{'} (z) = \frac{(6 z^{2} - 2) (z + 1) - (2 z^{3} - 2 z + 1) (1)}{(z + 1)^{2}} = \frac{4 z^{3} + 6 z^{2} - 3}{(z + 1)^{2}}

Now, plug in $z = 1$ :

B = Φ^{'} (1) = \frac{4 (1)^{3} + 6 (1)^{2} - 3}{(1 + 1)^{2}} = \frac{4 + 6 - 3}{4} = \frac{7}{4}

$B = 7 / 4$

To find $C$ :

First, find $Φ^{″} (z)$ by differentiating $Φ^{'} (z) = \frac{4 z^{3} + 6 z^{2} - 3}{(z + 1)^{2}}$ :

Φ^{″} (z) = \frac{(12 z^{2} + 12 z) (z + 1)^{2} - (4 z^{3} + 6 z^{2} - 3) (2 (z + 1))}{(z + 1)^{4}}

Simplify by canceling a $(z + 1)$ from all terms:

Φ^{″} (z) = \frac{(12 z^{2} + 12 z) (z + 1) - 2 (4 z^{3} + 6 z^{2} - 3)}{(z + 1)^{3}}

Φ^{″} (z) = \frac{(12 z^{3} + 24 z^{2} + 12 z) - (8 z^{3} + 12 z^{2} - 6)}{(z + 1)^{3}} = \frac{4 z^{3} + 12 z^{2} + 12 z + 6}{(z + 1)^{3}}

Now, plug in $z = 1$ :

Φ^{″} (1) = \frac{4 + 12 + 12 + 6}{(1 + 1)^{3}} = \frac{34}{8} = \frac{17}{4}

Finally, remember $C = Φ^{″} (1) / 2!$ :

C = \frac{17 / 4}{2} = \frac{17}{8}

$C = 17 / 8$

4. Final Result

Assemble all the pieces:

$G (z) = 1$
$A = 1 / 2$
$B = 7 / 4$
$C = 17 / 8$
$D = - 1 / 8$

\frac{z^{4}}{(z - 1)^{3} (z + 1)} = 1 + \frac{1 / 2}{(z - 1)^{3}} + \frac{7 / 4}{(z - 1)^{2}} + \frac{17 / 8}{z - 1} - \frac{1 / 8}{z + 1}

R (z) = \underset{G (z)}{\underset{⏟}{1}} + \underset{Pole 1}{\underset{⏟}{[\frac{1 / 2}{(z - 1)^{3}} + \frac{7 / 4}{(z - 1)^{2}} + \frac{17 / 8}{z - 1}]}} + \underset{Pole 2}{\underset{⏟}{[\frac{- 1 / 8}{z + 1}]}}