# Zeros in the complex plane and convergence

I'm doing some number theory which requires some work in $\mathbb{C}$, but unfortunately my complex analysis is a little rusty.

A text I am reading states the following:

...and given that $\sum_{\rho}|\rho|^{-2}$ converges, it follows that the product $\prod \limits_\rho (1-\frac{z}{\rho})e^{z/\rho}$ converges to an entire function with zeros at $z=\rho$ and nowhere else.

Could anyone explain, preferably without going into too much detail (just refreshing my memory!) why this is? Here, $\rho$ ranges over the set of zeros of an integral function $f: \mathbb{C} \to \mathbb{C}$ of order 1, which is not identically zero. It's obvious that the product is zero at each $\rho$; what do we need to know it converges to an entire function with zeros nowhere else?

For convergence, do we simply need to bound the modulus above by some finite value at each $z$? And since it's an infinite product, how do we confirm that it's zero nowhere else; a lower bound of the same manner? Finally, what tells us it is entire? Again, given that it's an infinite product I don't know how trivial that claim is; not very hard I'm sure but perhaps requiring some small justification.

As I said, I have probably learned all this at some point in my life so no need for extraordinary detail, just a brief explanation would be very helpful. It all seems vaguely reminiscent of a course I took on Riemann surfaces a few years back. Thanks!

• Doesn't this follow from the existence part of the Weierstrass factorization theorem? – user17762 Jun 7 '12 at 18:31
• @Marvis I don't think so. Because the canonical factor are all of the form $E_1(x) = (1 - x)e^{x}$. In the Wierstrass Product, you have $\prod_{n \in \omega}E_{n}(\frac{x}{a_n})$ where $a_n$ are the roots. Note that the canonical factors keep going up. However, maybe this follows from Hadamard if you can show the convergence condition implies order of growth with integer part 1. – William Jun 7 '12 at 18:34
• If it seems like this is wrong without further context then I can provide more information about the proof this statement came from: I had hoped that the information I provided was sufficient but perhaps I was wrong, apologies if so. – Spyam Jun 7 '12 at 18:37
• You want a version of the following lemma: let $a_n$ be a sequence such that $\sum |a_n|^2$ converges. Then $\prod (1 + a_n)$ converges if and only if $\sum a_n$ converges. The proof proceeds by taking logarithms. – Qiaochu Yuan Jun 7 '12 at 18:46
• Taking logs you can turn the product into a sum and have $$\sum_\rho ( \log(1-z/\rho) + z/\rho) \approx \sum_\rho (-z/\rho - 1/2 (z/\rho)^2 + z/\rho) =-z^2/2 \sum_\rho \rho^{-2}$$ Where everything needs to be made precise, but the above line is certainly the rough idea. This product should converge uniformly on compact sets and this will show that the limit is entire, and that there will be no new zeros popping up in the limiting process (that's a general property of uniform limits of holomorphic functions). – Sam Jun 7 '12 at 19:31

Just for ease of notation, enumerate the zeros as $(\rho_n)$. Then for any $R>0$ there exists $n_R$ such that $|\rho_n| > 2R$ for $n\ge n_R$. For any $N > n_R$, the function $$g_N(z) = \prod_{n=n_R}^N \left(1-\frac{z}{\rho_n}\right) e^{z/\rho_n}$$ has no roots in $|z|<R$, so there exists an analytic branch of $$h_N(z) = \log g_N(z) = \sum_{n=n_R}^N \left[ \log\left(1-\frac{z}{\rho_n}\right) + \frac{z}{\rho_n}\right]$$ where we choose the principal branch of the logarithm for each term in the sum. Using the fact that $\left|\frac{z}{\rho_n}\right|<\frac12$ for $|z|<R$ and $n\ge n_R$, and the fact that there exists a constant $C$ such that for $|w| <\frac12$ we have $|\log(1-w)+w| \le C|w|^2$, again for the principal branch of the logarithm (which follows easily from the power series expansion), we get that $$\left| \log\left(1-\frac{z}{\rho_n}\right) + \frac{z}{\rho_n}\right| \le C \frac{|z|^2}{|\rho_n|^2} \le \frac{CR^2}{|\rho_n|^2}$$ for all $n \ge n_R$ and $|z|<R$. By assumption the series over those terms converges, so $$\lim_{N\to\infty} h_N(z) = \sum_{n=n_R}^\infty \left[ \log\left(1-\frac{z}{\rho_n}\right) + \frac{z}{\rho_n}\right]$$ converges uniformly on $|z|<R$ to an analytic limit $h(z)$. This implies that $$\lim_{N\to\infty} g_N(z) = \lim_{N\to\infty} e^{h_N(z)} = e^{h(z)} = \prod_{n=n_R}^\infty \left(1-\frac{z}{\rho_n}\right) e^{z/\rho_n}$$ is a non-zero analytic function on $|z|<R$. This shows that $$f(z) = \prod_{n=1}^\infty \left(1-\frac{z}{\rho_n}\right) e^{z/\rho_n}$$ converges uniformly to an analytic function on $|z|<R$, whose only zeros are $z=\rho_n$ with $n<n_R$ and $|\rho_n|<R$. Since $R$ was arbitrary to begin with, the claim follows, and the convergence of the infinite product is locally uniform in the whole plane.