Let $f$ have a pole of order $m$ at $z_0$ and let $g$ have a pole of order $n$ at $z_0$. Classify the isolated singularity of $f/g$ at $z_0$.

Question

Please help me understand and how to start. Let $f$ have a pole of order $m$ at $z_0$ and let $g$ have a pole of order $n$ at $z_0$. Classify the isolated singularity of $\displaystyle{f/g}$ at $z_0$.

Answer 1

Since $f$ has a pole of order $m$ at $z_0$, $f$ can be written on the form

$$f(z)=\frac{\phi(z)}{(z-z_0)^m},$$

for an analytic function $\phi$ in a neighborhood around $z_0$. Correspondingly, $g$ may be written as

$$g(z)=\frac{\psi(z)}{(z-z_0)^n},$$

for an analytic function $\psi$ in a neighborhood around $z_0$. Thus,

$$\frac{f(z)}{g(z)}=\frac{\phi(z)}{(z-z_0)^m} \cdot \frac{(z-z_0)^n}{\psi(z)}= \frac{\phi(z)}{\psi(z)} (z-z_0)^{n-m}.$$

Now you see that if $n \geq m$, $\frac{f}{g}$ has a removable singularity at $z_0$. If $m > n$ however, we have a pole of order $m-n$.