If an exponential function can be defined in a field $\mathbb F$ as: $$ \exp(z)=\sum_{n=0}^\infty \frac{z^n}{n!} \qquad z \in \mathbb{F} \qquad (1) $$ we can prove (using the binomial formula and the Cauchy rule for the product of series) that $$ \exp(x+y)=\exp(x)\exp(y)\qquad (2) $$ and, if in the field there is an element $\tau \ne 0$ such that $\exp(\tau)=1$ than we see that the exp function is periodic with period $\tau$ because: $$ \exp(z+\tau)=\exp(z)\exp(\tau)=\exp(z) $$ and we can have also an element $\sigma= \tau/2$ such that $\exp(\sigma)=-1$ and an element $\iota=\sigma/2$ such that $[\exp(\iota)]^2=-1$
The classical example is $\mathbb{F}=\mathbb{C}$ where $\tau=2i\pi$ and $\exp(\iota)=i$.
Now, if we use $(1)$ to define the exp function in a non commutative ring, the property $(2)$ is not valid in general, and we can have different elements whose square is $-1$, as in the quaternion ring $\mathbb{H}$ where $\mathbf i^2=\mathbf j^2 =\mathbf k^2=-1$.
What can we say in this case about the periodicity of the exp function?
This question is connected to : Periodicity of the exponential function in a field , that has no answer.