3.8 Zeros and poles

Theorem 3.40

Suppose \(p\) and \(q\) are analytic at \(z_0\in \mathbb {C}\) with \(p(z_0)\neq 0\). Suppose further that \(q\) has a zero or order \(m{\lt}\infty \) at \(z_0\). Then \(p(z)/q(z)\) has a pole of order \(m\) at \(z_0\).

Proof ▼

In some neighborhood of \(z_0\), we can write \(q(z) = (z-z_0)^m g(z)\) where \(g(z_0)\neq 0\). Then \(h(z) = p(z)/g(z)\) is analytic at \(z_0\) and \(h(z_0)\neq 0\). And \(p(z)/q(z) = h(z)/(z-z_0)^m\). So \(p(z)/q(z)\) has a pole of order \(m\) at \(z_0\).

Theorem 3.41

Suppose \(p\) and \(q\) are analytic at \(z_0\), and suppose further that:

\(p(z_0)q'(z_0)\neq 0\).
\(q(z_0) = 0\).

Then \(z_0\) is a simple pole of \(p/q\) and

\[ \operatorname*{\mathrm{Res}}_{z=z_0} \frac{p}{q} = \frac{p(z_0)}{q'(z_0)}. \]

Proof ▼

We can write \(q(z) = (z-z_0)^m g(z)\) with \(g\) analytic at \(z_0\), \(g(z_0)\neq 0\) and \(m {\gt} 0\). Then \(q'(z_0) = m(z_0-z_0)^{m-1} g(z_0) = 0\) unless \(m = 1\). So, by a, we must have \(m = 1\), and, therefore, \(q'(z_0) = g(z_0)\).

Then, set \(h(z) = p(z)/g(z)\). We get that \(h(z_0) \neq 0\), an consequently,

\begin{align*} \operatorname*{\mathrm{Res}}_{z=z_0} \frac{p}{q} & = \operatorname*{\mathrm{Res}}_{z=z_0} \frac{h}{z-z_0} = h(z_0)\\ & = \frac{p(z_0)}{g(z_0)} = \frac{p(z_0)}{q'(z_0)} \end{align*}

as desired.