You really don't have to go too much into depth in choosing your branch of logarithm. The principal branch is good enough if you add a \( \rho \) function.
If you write,
\(
\text{Tom}(s) = \Omega_{j=1}^\infty e^{A(s-j)z}\,\bullet z\\
\)
And,
\(
\text{Tom}_A(s) = A(s)\text{Tom}(s) = \log \text{Tom}(s+1)\\
\)
And construct a sequence \( \rho_n(s) \) where,
\(
\rho^{n+1}(s) = \log(1+\frac{\rho^n(s+1)}{\text{Tom}_A(s+1)}) + \log A(s+1)\\
\)
where \( \frac{\rho^n(s+1)}{\text{Tom}_A(s+1)} \) is very small for large \( \Re s \). So you are effectively calculating a \( \log(1+\Delta) \) for \( \Delta \) small, as opposed to a \( \log(X) \) where \( X \) is large. The branching won't be an issue at all.
Where then,
\(
\text{tet}_{\text{Tom}}(s + x_0) = \text{Tom}_A(s) + \rho(s)\\
\)
Remember Tommy that,
\(
\lim_{\Re(s) \to \infty} \text{Tom}_A(s) \to \infty\\
\)
Even though there are dips to zero this is still the asymptotic behaviour.
If you write,
\(
\text{Tom}(s) = \Omega_{j=1}^\infty e^{A(s-j)z}\,\bullet z\\
\)
And,
\(
\text{Tom}_A(s) = A(s)\text{Tom}(s) = \log \text{Tom}(s+1)\\
\)
And construct a sequence \( \rho_n(s) \) where,
\(
\rho^{n+1}(s) = \log(1+\frac{\rho^n(s+1)}{\text{Tom}_A(s+1)}) + \log A(s+1)\\
\)
where \( \frac{\rho^n(s+1)}{\text{Tom}_A(s+1)} \) is very small for large \( \Re s \). So you are effectively calculating a \( \log(1+\Delta) \) for \( \Delta \) small, as opposed to a \( \log(X) \) where \( X \) is large. The branching won't be an issue at all.
Where then,
\(
\text{tet}_{\text{Tom}}(s + x_0) = \text{Tom}_A(s) + \rho(s)\\
\)
Remember Tommy that,
\(
\lim_{\Re(s) \to \infty} \text{Tom}_A(s) \to \infty\\
\)
Even though there are dips to zero this is still the asymptotic behaviour.