# $\frac{f(z)}{(z - z_1)…(z - z_N)}$ is bounded near each $z_j?$

In a book I am reading, it states the title where $f$ is a holomorphic function in an open set $\Omega$ which contains the closure of a disc centered at the origin, $f(0) \neq 0,$ $f$ vanishes nowhere on the boundary of the disc, and $z_1, ..., z_N$ are the zeros of $f$ inside the disc. How is $g(z) = \frac{f(z)}{(z - z_1)...(z - z_N)}$ bounded near each $z_j?$ Isn't it the opposite? Wouldn't $g$ blow up near each of the zeros? The book uses this to claim that $z_1, ..., z_N$ are removable singularities - which I understand. But not the bounded argument.

Since $z_1$ is a zero of $f$, there is $m \in \mathbb N$ and a holomorphic function $h$ on $\Omega$ such that
$f(z)=(z-z_1)^mh(z)$ for all $z \in \Omega$ and $h(z_1) \ne 0$. Then
$g(z) = \frac{(z-z_1)^{m-1}h(z)}{(z - z_2)...(z - z_N)}$
This shows that $g$ is bounded near $z_1$
Similar arguments are valid for $z_j, j \ge 2$.