Complex Variables Series Coefficient proof. I was studying for my Complex Analysis(an elementary undergraduate Complex Analysis course) exam and came across this question in the textbook.  I know that in order for there to be a pole of some order, the limit of f(z) would have to go to infinity.  I also know that if f(z) has a removable singularity then it is bounded.  Thus, I am not sure how if two functions, each with poles of different orders, has a product with a removable singularity.  Any advice, tips, and suggestions would be much welcomed.  I have attached this question as an image.  Please note, that this is for an elementary Complex Analysis course, so please nothing too advanced, I am really trying to understand the concepts.  
 A: Since $f$ has a zero of order $m$ at $z_0$, its Taylor series in the disk $|z - z_0 | < R$ looks like:
$$ f(z) = a_m(z - z_0)^m + a_{m+1} (z - z_0)^{m+1} + a_{m+2}(z - z_0)^{m + 2} + \dots $$
Therefore, you can write
$$ f(z) = (z - z_0)^m p(z),$$
where
$$ p(z) = a_m + a_{m+1}(z - z_0) + a_{m+2}(z - z_0)^2 + \dots$$
Since $p(z)$ is given by a power series, and since all power series are holomorphic within their disk of convergence, it follows that $p(z)$ is holomorphic on the disk $| z - z_0 | < R$.
Similarly, since $g$ has a pole of order $l$ at $z_0$, its Laurent series on the punctured disk $0 < | z - z_0 | < R$ looks like:
$$ g(z) = b_{-l}(z - z_0)^{-l} + b_{-l+1}(z - z_0)^{-l+1} + b_{-l+2}(z - z_0)^{-l+2} + \dots.$$
So you can write
$$ g(z) = (z - z_0)^{-l}q(z),$$
where
$$ q(z) = b_{-l} + b_{-l+1}(z - z_0) + b_{-l+2}(z - z_0)^2 + \dots$$
Since $q(z)$ is given as a power series, $q(z)$ is holomorphic on the disk $| z - z_0 | < R$; in particular, it is well-defined and holomorphic at the point $z = z_0$.
Finally,
$$ f(z)g(z) = (z - z_0)^{m-l} p(z) q(z).$$
Since $m - l \geq 0$, and since $p(z)$ and $q(z)$ are both holomorphic on $| z - z_0 | < R$ (including at the point $z = z_0$), we see that the right-hand side is holomorphic on the disk $|z - z_0| < R$ (including at the point $z = z_0$). Thus we have succeeded in extending $f(z)g(z)$ to a holomorphic function on $| z - z_0 | < R$ (including at the point $z = z_0$), which proves that $f(z)g(z)$ has a removable singularity at $z = z_0$.
