Proof that $\zeta (z)=\frac{\sigma '(z)}{\sigma (z)}$

Question

The Weierstrass $\sigma$ function of a lattice $\Omega$ is defined by $$\sigma (z)=z\prod_{\omega\in \Omega\setminus\{0\}}\left(1-\frac{z}{\omega}\right)\exp\left(\frac{z}{\omega}+\frac{1}{2}\frac{z^2}{\omega^2}\right).$$

The Weierstrass $\zeta$ function of a lattice $\Omega$ is defined by $$\zeta(z)=\frac{1}{z}+\sum_{\omega\in\Omega\setminus\{0\}}\left(\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega^2}\right).$$

I want to prove that $$\zeta (z)=\frac{\sigma'(z)}{\sigma(z)}.$$

I've seen the following proof of the above theorem:

$$\frac{d}{dz}\log\sigma (z)=\frac{d}{dz}\log z+\sum_{\omega\in\Omega\setminus\{0\}}\frac{d}{dz}\left(\log\left(1-\frac{z}{\omega}\right)+\frac{z}{\omega}+\frac{1}{2}\frac{z^2}{\omega^2}\right),$$ from which the identity follows.

The complex logarithm is defined by $$\log z=\log |z|+i\operatorname{arg}z$$ where $\operatorname{arg}$ is the complex argument with the nonpositive real axis as a branch cut; $\operatorname{arg}z\in (-\pi,\pi]$.

Now I see $5$ problems with the above "proof":

1. $\log$ is not differentiable everywhere because of the branch cut.
2. $\log zw=\log z+\log w$ if $-\pi\lt \operatorname{arg}z+\operatorname{arg}w\le \pi$; the identity can fail outside this region.
3. $\log zw=\log z+\log w$ does not imply $\log\prod=\sum\log$ where $\prod$ is an infinite product.
4. $\log e^z=z$ for $z$ such that $-\pi\lt \operatorname{Im}z\le \pi$; the identity can fail outside this region.
5. Interchanging the derivative and infinite series could be justified by uniform convergence, but how would one prove uniform convergence in this case?

Can the above problems be patched? If so, how? Alternatively, can the theorem in question be proved without using logarithms?

I thought that $$\mathrm{LHS}(z)=\mathrm{RHS}(z)+2\pi ik$$ and then differentiating could work, but then I realized that $k$ depends on $z$ and has jump discontinuities, so again we run into the problem of non-existent derivatives.

Answer 1

Idea. You can think backward. Instead of taking logarithm to $\sigma(z)$, you can first create a logarithmic version of $\sigma(z)$ and then work with this. That is, we first realize a version of

$$ \text{“} \ \log\sigma(z) = \log z + \sum_{\omega\in \Omega\setminus\{0\}} \left( \log \left(1-\frac{z}{\omega}\right) + \frac{z}{\omega} + \frac{z^2}{2\omega^2} \right) \ \text{”} $$

by (1) localizing the range of $z$ onto sufficiently small and nice domain, and (2) choosing an appropriate version for each of the complex logarithms appearing in the right-hand side. This allows to fill the gap of the proposed argument and make it rigorous.

Implementation. To implement this idea, we need some preparation:

• To verify $\frac{\sigma'(z)}{\sigma(z)} = \zeta(z)$ on $\mathbb{C}\setminus\Omega$, it suffices to verify this on every open ball in $\mathbb{C}\setminus\Omega$.

• Let $U$ be any open ball contained in $\mathbb{C}\setminus\Omega$.

• Fix $R > 0$ so that $U$ is contained in the disk of radius $R$ centered at the origin.

• Define $L_{0} : U \to \mathbb{C}$ as a holomorphic function so that $\exp(L_{0} (z)) = z$. This is possible because $U$ is simply connected and $z$ has neither poles nor zeros on $U$. Also, there are infinitely many choices of $L_0$, but any choice will work.

• For each $\omega \in \Omega\setminus\{0\}$, define $L_{\omega} : U \to \mathbb{C}$ as a holomorphic function satisfying the identity $\exp(L_{\omega} (z)) = 1 - \frac{z}{\omega}$. This is possible by the similar reason as in the previous step.

• Finally, define $\texttt{log_sigma} : U \to \mathbb{C}$ by

  \begin{align*} \texttt{log_sigma}(z) &= L_0(z) + \sum_{\substack{w \in \Omega\setminus\{0\} \\ |w| < 2R }} \left( L_{\omega}(z) + \frac{z}{\omega} + \frac{z^2}{2\omega^2} \right) \\ &\hspace{5em} + \sum_{\substack{w \in \Omega\setminus\{0\} \\ |w| \geq 2R }} \left( \log\left(1 - \frac{z}{\omega}\right) + \frac{z}{\omega} + \frac{z^2}{2\omega^2} \right). \end{align*}

The defining sum for $\texttt{log_sigma}$ converges uniformly in $U$, hence it defines a holomorphic function over $U$ and term-wise differentiation is applicable. Moreover, it is clear that

$$ \exp(\texttt{log_sigma}(z)) = \sigma(z). $$

Then by using $L_0'(z) = \frac{[\exp(L_0(z))]'}{\exp(L_0(z))} = \frac{1}{z}$ and $L_{\omega}'(z) = \frac{[\exp(L_{\omega}(z))]'}{\exp(L_{\omega}(z))} = \frac{1}{z-\omega}$, we get

$$ \frac{\sigma'(z)}{\sigma(z)} = \texttt{log_sigma}'(z) = \frac{1}{z} + \sum_{w \in \Omega\setminus\{0\}} \left( \frac{1}{z-\omega} + \frac{1}{\omega} + \frac{z}{\omega^2} \right) = \zeta(z). $$