If $f$ and $g$ have poles of order $m,n$ respectively, what is the pole order of $f/g$?

If $f$ has a pole of order $m$ at $z_0$ and $g$ has a pole of order $n$ at $z_0$ then there exist analytic functions $F$ and $G$ in some disc around $z_0$ such that $f(z)=(z-z_0)^{m}F(z)$, $g(z)=(z-z_0)^{n}G(z)$ and $F(z) \neq 0, G(z) \neq 0$ in that disc. Hence $\frac {f(z)} {g(z)}=(z-z_0)^{m-n} H$ where $H=\frac F G$. Note that $H$ is analytic in the disc and does not vanish in it. What conclusions can you draw from this?

Since$$\frac fg(z)=(z-z_0)^{n-m}\times\frac{c_{-m}+c_{-m+1}(z-z_0)+\cdots}{d_{-n}+d_{-n+1}(z-z_0)+\cdots}$$then $z_0$ is:

• a zero of order $n-m$ if $n>m$;
• a removable singularity if $n=m$;
• a polo of order $m-n$ if $n<m$.

If $f$ has a pole at $z_0$ of order $m$, then there is a function $h$ which is holomorphic in a neighborhood of $z_0$ such that

$f(z)= \frac{h(z)}{(z-z_0)^m}$ and $h(z_0) \ne 0.$

If $g$ has a pole at $z_0$ of order $n$, then there is a function $k$ which is holomorphic in a neighborhood of $z_0$ such that

$f(z)= \frac{k(z)}{(z-z_0)^n}$ and $h(z_0) \ne 0.$

Then we have

$$\frac{f(z)}{g(z)}=\frac{h(z)}{k(z)}(z-z_0)^{n-m}.$$

If $n \ge m$, then $f/g$ has a removable singularity at $z_0).

If $n<m$, the $f/g$ has a pole of order $m-n.$