Function on $\mathbb{C}$ with all primes as zeros? According to the Weierstraß factorization theorem, an entire function with all primes as zeros would be (if I didn't mess up):
$$\tilde P(z) = \prod_{p \text{ is prime}} \left(1-\frac{z}{p}\right) \cdot e^{z/p} $$ for $z\in\mathbb{C}$.
Formally I can define the slightly different function
$$P(z) = \prod_{p \text{ is prime}} \left(1-\frac{z}{p}\right)$$ 
for $z\in\mathbb{C}$.
Question: For which $z\in\mathbb{C}$ does the latter converge apart from the obvious $z=0$ or $z$ is prime?
Thoughts:


*

*Taking the $\log$ this translates to the question for which $z$ $$\sum_{p \text{ is prime}}\log(1-z/p)$$ converges. I am guessing that this is not the case because thinning the harmonic series out to just primes, $\sum 1/p$, does not make it convergent and since $\sum \log(1-1/n)$ is not convergent, thinning out to primes likely does not help either.

*Yet for $\tilde P$ the $\log$ yields $$\sum_{p \text{ is prime}}(z/p+\log(1-z/p)),$$ which is, in absolute value, even larger but should converge because $\tilde P(z)$ converges. Hmm???

 A: Remark: There are two slightly different definitions of convergence of an infinite product $\prod_{n=1}^\infty a_n$. One definition is that all factors $a_n$ are nonzero and that $\lim_{N \to \infty} \prod_{n=1}^N$ exists and it not zero.
The other definition allows that finitely many factors are zero, and  requires that for some $n_0$, $\lim_{N \to \infty} \prod_{n=n_0}^N$ exists and it not zero. Using this definition, the value of a convergent infinite product is zero if and only if one factor is zero.
With the first definition, both of your products are divergent if $z$ is a prime.
I'll use the second definition here, which has the advantage that the case “$z$ is a prime” does not need to be considered separately.

It follows e.g. from the Taylor series of $\log(1+w)$ that
$$ 
|\log(1+w) - w | < K |w|^2
$$
for $|w| < 1/2$ and some constant $K > 0$. Therefore, for $|z| < R$ and all primes $p > 2|z|$,
$$ \tag{*}
\left| \log\left(1-\frac{z}{p}\right) + \frac{z}{p}\right| < \frac{KR^2}{p^2} \, .
$$
This implies the (locally uniform) convergence of $\sum\limits_{p > 2|z| \text{ is prime}} \log\left(1-\frac{z}{p}\right) + \frac{z}{p}$ and thus the convergence of the infinite product $$\prod_{p \text{ is prime}} \left(1-\frac{z}{p}\right) \cdot e^{z/p}\, .$$ 
For non-zero $z$ is the series $\sum\limits_{p \text{ is prime}} \frac{z}{p}$ divergent, so that $(*)$ also implies the divergence of $\sum\limits_{p > 2|z| \text{ is prime}} \log\left(1-\frac{z}{p}\right)$ and consequently the divergence of
$$\prod_{p \text{ is prime}} \left(1-\frac{z}{p}\right)\, .$$ 
