Is it possible to plug hypercomplex numbers into the Riemann Zeta function?

Question

I'm aware of the more detailed question:

How to raise a number to a quaternion power

However, from a more high-level perspective (read: probably less mathematical, hence my choice for raising this here.) I'm trying to understand this in terms of the Riemann Zeta function and [multivalued functions][2].

From what I understand of the answer given in the question mentioned above it does not make sense to raise a real number to a hypercomplex (quaternion or octonion) power because it produces multiple values.

Doesn't this also apply to complex number exponents?

Isn't there a way to plug hypercomplex numbers into the Riemann Zeta function?

Answer 1

Let $A$ be a normed finite-dimensional $\mathbb{R}$-algebra, so $A$ is isomorphic to a sub-algebra of $\mathbb{R}^{m \times m}$.

$\exp(s) = \sum_{k=0}^\infty \frac{s^k}{k!}$ converges for every $s \in A$.

Fix some $s \in A$. From the Jordan normal form we obtain $s = p (\Lambda+b) p^{-1}$ where $\Lambda$ is a complex diagonal matrix, $b^m = 0$ and $\Lambda,b$ commute. Thus $$n^{-s} = \exp(-s \log n) =\exp(-p(\Lambda+b)p^{-1} \log n)=p\exp(-(\Lambda+b) \log n)p^{-1}\\=p\exp(- \Lambda\log n)\exp(-b \log n)p^{-1} = p n^{-\Lambda} \sum_{k=0}^m \frac{(- \log n)^k}{k!}b^k p^{-1}$$ where $n^{-\Lambda}$ is the complex diagonal matrix with entries $n^{-\Lambda}_{jj}=n^{-\Lambda_{jj}}$. Therefore $$\zeta(s) = \sum_{n=1}^\infty n^{-s} =p \sum_{k=0}^m \frac{b^k}{k!} \zeta^{(k)}(\Lambda) p^{-1}$$ where $\zeta^{(k)}(\Lambda)$ is the complex diagonal matrix with entries $\zeta^{(k)}(\Lambda)_{jj}=\zeta^{(k)}(\Lambda_{jj})$.

For $A$ a general normed $\mathbb{R}$-algebra, something similar holds at least when $s$ is compact.

Therefore all the formulas about $\zeta$ stay true in this setting and up to some trivial terms, it reduces to the usual function $\zeta(z),z \in \mathbb{C}$ and its $k$-th derivatives.

Answer 2

In tessarines the full answer is

$$\zeta ((a+b i)+j (c+d i))=\frac{1}{2} (\zeta (a+c+b i+d i)+\zeta (a-c-d i+b i))+\frac{1}{2} j (\zeta (a+c+b i+d i)-\zeta (a-c-d i+b i))$$

The bicomplex numbers are isomorphic to tessarins, so tessarine $j$ is equal to $-k$ in bicomplex numbers. This is a general formula that can be used for any function defined on complex numbers.