I would appreciate help in how to show that
On a compact subset of {$z \in \mathbb{C}: 1/2 < \Re (z) < 1$}:
"Given a holomorphic function with an isolated zero, any "nearby" holomorphic function will also have a zero."
This statement is from an article on Zeta Function Universality, http://en.wikipedia.org/wiki/Zeta_function_universality.
I would presume that "nearby" implies that if $z_0$ is a zero of $f$ and you represent $f$ as a power series around $z_0$, then you can also represent the "nearby" $g$ also as a power series about $z_0$.
Thus you would have
$$f(z) =\sum_{n=0}^\infty a_n(z-z_0)^n = (z - z_0)^k\sum_{n=k + 1}^\infty a_n(z-z_0)^n$$
where $f$ has a zero of order $k$ at $z_0$, and $a_{k + 1}$ is the first coefficient $\neq 0$.
Then the assertion would imply that you can similarly represent the "nearby" $g$ by
$$g(z) =\sum_{n=0}^\infty b_n(z-z_0)^n = (z - z_0)^l\sum_{n=l + 1}^\infty b_n(z-z_0)^n$$
Assuming this is correct so far. I would appreciate help proving the statement.
Thanks very much.
EDIT: This is the relevant part of the article:
"Note that the function f is not allowed to have any zeros on U. This is an important restriction; if you start with a holomorphic function with an isolated zero, then any "nearby" holomorphic function will also have a zero. According to the Riemann hypothesis, the Riemann zeta function does not have any zeros in the considered strip, and so it couldn't possibly approximate such a function."