0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $z_j, j \ge 2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.