2.4Rational Functions

A rational function $R(z)$ is the quotient of two polynomials:

For simplicity, suppose $R(z)$ is simplified fully, and so $P$ and $Q$ share no common factors. We will say $R(z) = \infty$ at the zeros of $Q(z)$, where $\infty \in \hat{\bC}$. The zeros at $Q(z)$ are referred to as the poles of $R(z)$, and the order of a pole is defined to be the order/multiplicity of the root of $Q(z)$.

Note that, for $R(z) = P(z)/Q(z)$, we have that

Note that if $z_0$ is a pole of $R$, then $z_0$ is also a pole for $R'$—we seek to understand how differentiation affects the order of a pole. Write $Q(z) = (z - z_0)^nA(z)$, where $n$ is the order of $R(z_0)$. Then note

where $V(z)$ is the residual polynomial such that $V(z_0) \ne 0$. Thus $R'(z_0)$ has order $n + 1$.

We note the value of a rational function at a pole is at $\infty \in \hat{\bC}$. But what about the behavior of a rational function at $\infty$? That is, how do we determine if a rational function $R(z)$ has a pole at $\infty$?

We can let $z \to \infty$ and observe $R(z)$, but this wouldn't determine the order of a pole at $\infty$. Instead, we define $R_1(z) = R(1/z)$, and observe what happens to $R_1$ as $z \to 0$. This is perfect to determine the behavior of $R$ at $\infty$.

For some arbitrary rational function, note

Accordingly, if $m > n$, then $R(\infty) = 0$ with order $m - n$. If $m < n$, then $R(\infty)$ has a pole of order $n - m$. If $m = n$, then $R(\infty) = a_n / b_m$.

Again, write $R(z) = P(z)/Q(z)$ in its presentation as in the last remark, where $\deg(P) = n$ and $\deg(Q) = m$. In the finite plane $\bC$, we note $R$ has $n$ zeros and $m$ poles.

Now let's consider this in the one-point compactification of the plane, $\hat{\bC} = \bC \cup \{\infty\}$. If $m > n$, we have a zero at $\infty$ of order $m - n$; if $m < n$, we have a pole of order $n - m$; if $m = n$, we have neither. In any case, it follows that the number of zeros and the number of poles of $R$, as considered in the infinite plane, coincide and are equivalent to $\max(m, n)$. We say that the order of a rational function is the number of zeros/poles it has in the infinite plane.

Note that if $R(z)$ has order $p$, then $R(z) - a$ obviously also has order $p$.

We seek to show that any rational function $R(z) = P(z)/Q(z)$ has a partial fraction decomposition. Partial fraction decomposition is a method to encode the behavior of $R$ at its poles.

To start, suppose $R$ has a pole at $\infty$. Then $\deg(P) > \deg(Q)$, and so by long division, we have that

where $c_0 = G(0)$. What we have done is split $R(z)$ as the sum of two parts: one part that blows up at $\infty$ (this encodes the behavior of $R$ at $\infty$), and one part that is finite at $\infty$. After normalizing $G$ so that $G(0) = 0$, we say $G$ is the singular part of $R(z)$ at $\infty$, whereas $H$ is the finite part at $\infty$. This terminology applies to other poles, as we will soon see.

Now let $\lambda_1, \dots, \lambda_n$ denote the distinct, finite poles of $R$. To create a decomposition similar to the one at $\infty$, we use a transformation that sends $\lambda_k$ to $\infty$. $\zeta = (z - \lambda_k)^{-1}$ is the appropriate transformation, since when writing $z = \lambda_k + \zeta^{-1}$, the point $\lambda_k$ corresponds to a pole at $\infty$. We now apply this type of decomposition

where $G_k$ is the singular part of $R(z)$ at $\lambda_k$ and $H_k$ is the finite part.

Finally, we seek to show

Define

We will first show $f$ has no poles. We notice immediately the only candidates at which $f$ has a pole are $\lambda_1, \dots, \lambda_n, \infty$.

We first observe how $f$ behaves at each $\lambda_k$. First, write

Note the first term reduces to $H_k((z - \lambda_k)^{-1})$, which is finite at the pole, and $G(z)$ and the sum do not diverge at $\lambda_k$. Thus $f$ does not have a pole at $\lambda_k$. Now, observe how $f$ behaves at $\infty$. Note $\lim_{z \to \infty} R(z) - G(z) = \lim_{z \to \infty} H(z)$, which is finite, and the other terms are also finite as $z \to \infty$. Thus $f$ has no pole at $\infty$ and thus has no poles at all.

The only rational functions that have no poles are identically constant. Absorbing this constant into $G(z)$, we have proven