Proposition

Suppose that $f$ has a zero of multiplicity $m$ at $a$. Explain why $1/f$ has a pole of order $m$ at $a$.

Solution

$f$ has a zero of multiplicity $m$ at $a$. By Theorem 8.14[A first course in complex analysis], this means that there exists a holomorphic function $g$ such that

• $g(a) \ne 0$,
• $f(z) = (z - a)^mg(z)$.

Moreover, there exists a disk $D[a, r]$ in which $a$ is the only zero of $f$. This implies that $g$ is never 0 in $D[a, r]$.

Then $f$ is holomorphic in the punctured disk $D’[a, r]$. Therefore,

for all $z \in D’[a, r]$.

By Corollary 9.6[A first course in complex analysis], $1/f$ has a pole of order $m$ at $a$.