I've been reading about singularities, and I was wondering if is it possible to have a function (in the extended complex plane) that only has a pole of order $m$ at zero and a pole of order $n$ at infinity? If so, what would be the general form of such functions?
A function which has no singularities in the extended complex plane other than poles must be rational. Also, if a rational function is written in lowest terms as $f(z)=P(z)/Q(z)$, it will have a pole at every zero of $Q$. So, a function whose only singularities are poles at $0$ and infinity must be of the form $$f(z)=\frac{P(z)}{z^m},\qquad m\in {\Bbb Z}_{>0},\ \ P(z)\in {\Bbb C}[z].\ \ \ (*)$$ If $f$ is in lowest terms, we can also assume that $P(0)\ne 0$. $(*)$ will then clearly have a pole of order $m$ at $0$, and it will have a pole of order $n$ at infinity iff $P$ has degree $m+n$. Therefore, the answer to your question is that the general form of such a function is $$f(z)=\frac{P(z)}{z^m},\qquad m, n\in {\Bbb Z}_{>0}, \ P(z)\in {\Bbb C}[z],\ P(0)\ne 0,\ {\rm deg} P=m+n.$$ Another way of expressing such a function is by a finite Laurent series: $$f(z)=a_{-m} z^{-m} + a_{-(m-1)} z^{-(m-1)} + \cdots + a_{n-1} z^{n-1} + a_n z^n,$$ $$a_{-m}, \ldots, a_n\in {\Bbb C},\ \ a_{-m}\ne 0,\ a_n\ne 0.$$