# Cancellation of zeros and poles

Suppose that $f$ has a pole of order $m$ at $a$ and $g$ is holomorphic in $D(a;r)$ (open disc centered at a with radius $r>0$). Then at $a$ the function $fg$

(i) has pole of order $n-m$ if $g$ has zero of order $n$ ($n < m$) at $a$.

(ii) has removable singularity if $g$ has zero of order $n$ ($n > m$) at $a$.

Argument:

Since $f$ has a pole of order $m$, we can write it as $$f(z) = \lim_{z \to a} \frac{D}{(z-a)^m},$$ where $D$ is a non-zero constant.

Similarly, $$g(z) = \lim_{z \to a} {(z-a)^n}h(z),$$ where $h(z) \neq 0$.

Hence, \begin{align*} fg(z) &= \lim_{z \to a} \frac{D}{(z-a)^m} \times \lim_{z \to a} (z-a)^n h(z) \\ &= \lim_{z \to a} \frac{D}{(z-a)^m}{(z-a)^n}h(z). \end{align*}

For (i), since $m > n$, we have $(z-a)^{m-n}$ in the denominator, hence $fg$ has a pole of order $m-n$.

For (ii), since $m < n$, we have $(z-a)^{n-m}$ in the numerator, hence $fg$ should have zero of order $n-m$.

But why is there a removable singularity? Please explain.

• For what it's worth, your limit criteria are probably not what you mean. Normally, one says $f$ has a pole of order $n > 0$ at $a$ if there exists a holomorphic function $h$ such that $$f(z) = \frac{h(z)}{(z - a)^{n}},\qquad h(a) \neq 0,$$for all $z$ in some deleted open disk about $a$, and similarly for a zero of order $m$. Still, the idea of your calculation is correct. :) – Andrew D. Hwang Sep 20 '16 at 12:40

In case (ii), the product $fg$ isn't defined at $z=a$, since $f$ isn't.
Hence, at least formally, $fg$ has a singularity at $z=a$, but your computation shows that it is removable. (Putting $fg(a) = 0$ gives you a holomorphic function with a zero of order $n-m$ at $z=a$.)
• @SamChristopher in complex analysis, we consider the analytic continuation of functions, so (when it exists) $fg(a) = \lim_{z \to a} fg(z)$ – reuns Sep 20 '16 at 11:20
• @SamChristopher $f$ has a pole at $z=a$, so in particular $f$ is not defined at $a$. – mrf Sep 20 '16 at 13:09