Let $(\mathcal{A},\alpha)$ be a C* dynamical system, i.e. $\mathcal{A}$ is a unital C*-algebra and $\{\alpha_t\}_{t\in \mathbb{R}}$ a strongly continuous one-parameter group of *-automorphisms. For every $A\in \mathcal{A}$, the function $\alpha: \mathbb{R} \rightarrow\mathcal{A}, t\mapsto \alpha_t(A)$ can be analytically continued to a function $\alpha: \mathbb{C} \rightarrow\mathcal{A}, z\mapsto \alpha_z(A)$.
I want to show that the group property is conserved. More precisely I want to show that
$$\forall \mathbb{C}\ni z=a+i b : \alpha_{a+i b}(A) = \alpha_a(\alpha_{i b}(A))$$
Unfortunately, I don't know how show in general that a property is conserved during analytic continuation. How can one proceed in this particular example?
Let's start by assuming that such an analytic continuation exists.
In what follows, $z \in \mathbb C$ and $a, b, t \in \mathbb R$. The following are then $\mathcal A$-valued entire functions: $$ z \mapsto \alpha_z(A), \qquad z \mapsto \alpha_{a +z} (A). $$ The second of these is entire as shifting everything by a fixed constant $a$ will not affect the differentiability of the function. In other words, the composition of the entire maps $z \mapsto a +z$ and $z \mapsto \alpha_z(A)$ will also be entire.
We also have that $$ z \mapsto \alpha_z(\alpha_a(A)) $$ is entire. We are just applying the map to the fixed input $\alpha_a(A)$.)
These two entire maps agree on the real axis: $$ t \mapsto \alpha_{a +t}(A)=\alpha_t(\alpha_a(A)). $$
However two entire maps which agree on the real axis must be equal. We then have that $$ \alpha_{a+z}(A)=\alpha_z(\alpha_a(A)). $$ In particular, this implies that $$ \alpha_{a+ib}(A)=\alpha_a (\alpha_{ib}(A)). $$