Holomorphic Functional Calculus for Exponential Identity For a Banach algebra $A$ with unit $e$, then for $a\in A$, I want to prove $$\exp((\alpha+\beta)a)=\exp(\alpha a)\exp(\beta a)\qquad\text{for all }\alpha,\beta\in\mathbb{C}$$
So far I have said let $\Omega$ be an open set in $\mathbb{C}$ and $H(\Omega)$ the algebra of all complex holomorphic functions in $\Omega$. Define $A_{\Omega}:=\{a\in A|\sigma(a)\subset\Omega\}$, an open subset of $A$.
$\exp:\Omega\to\mathbb{C}$ is a holomorphic function thus $\exp\in H(\Omega)$.
Define $\widetilde{H}(A_{\Omega})$ as the set of $A$-valued functions $\widetilde{\exp}$, with domain $A_{\Omega}$, that arise from $\exp\in H(\Omega)$ by the formula $$\widetilde{\exp}(a)=\frac{1}{2\pi i}\int_{\Gamma}\frac{\exp(\lambda)}{\lambda e-a}d\lambda$$
$\widetilde{H}(A_{\Omega})$ is a complex algebra and $\exp\mapsto\widetilde{\exp}$ is an algebra homomorphism, in particular for all $a\in A_{\Omega}$.
But I cannot actually figure out how to use employ this theory to solve my question. In particular, I do not fully understand the mapping $\exp\mapsto\widetilde{\exp}$.
 A: The resolvent for $|\lambda| > \|x\|$ is
$$
       (\lambda e- x)^{-1}=\lambda^{-1}(e-\frac{1}{\lambda}x)^{-1}=\frac{1}{\lambda}\sum_{n=0}^{\infty}\frac{1}{\lambda^{n}}x^{n}=\sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}x^{n}.
$$
So when you integrate around a contour that encloses $|\lambda|\le \|x\|$, you get
\begin{align}
  \widetilde{\exp}(x) & =\frac{1}{2\pi i}\oint_{|\lambda|=\|x\|+\epsilon} e^{\lambda}\sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}x^{n}d\lambda \\
  & = \sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\oint \frac{e^{\lambda}}{\lambda^{n+1}}d\lambda\right)x^{n} \\
  & = \sum_{n=0}^{\infty}\frac{1}{n!}x^{n}.
\end{align}
So the functional calculus definition of $e^{x}$ agrees with the power series definition. What's nice, of course, is that the exponential properties for the functional calculus exponential definition are automatic, which proves that the properties hold for the power series definition, too.
A: You are trying to do this the hard way, through functional calculus. But $\exp$ on a Banach algebra can be defined more conviniently by a infinite series:
$$ \exp (a) = \sum_{n=0}^\infty \frac{1}{n!} a^n $$
Which is absolutely convergent for any $a$. Try to prove why. If you check proof of $ \exp (x+y)= \exp x \exp y$ for complex numbers, you can conclude that for $a,b$ in Banach algebra that commute the same holds true. Try to repeat the argument.

Edit: this definition and the one through functional calculus are equivalent, it follows from how we define $\exp$ for complex numbers.
