Prove that dual of a unital Banach algebra is nonempty using resolvent and Liouville's theorem

Question

Let $\mathcal{A}$ be a unital Banach algebra over complex numbers $\mathbb{C}$. For every $a \in \mathcal{A}$, let $\sigma(a)$ be the spectrum of $a$. Define the resolvent of $a$ to be $ R(a,z) = (z1_a - a)^{-1} $ which is defined for all z,a such that $z1_a - a$ is invertible. For each $\phi \in \mathcal{A}^*$, which is the dual of $\mathcal{A}$, $z \rightarrow \phi(R(a,z))$ is a $\mathbb C$-valued analytic bounded function. I believe my professor also said it is surjective. So by Liouville's theorem, it is constant and identically equals to zero (I am not familiar with the theorem, but I have no problem just accepting it). Finally, he mentioned that this gives us another proof (or a way to prove) of the existence of a large space of linear functionals, which unlike Hahn-Banach theorem, does not rely on the axiom of choice. I fail to see how that would work. Any suggestions/references?