Why is $z \mapsto E(e^zB_t)$ where $B_t$, $z\in \mathbb{C}$ is a Brownian motion is analytic.

Question

$z \mapsto E(e^zB_t)$ where $B_t$, $z\in \mathbb{C}$ is a Brownian motion is analytic. I am looking at the proof that for a one dimensional Brownian motion $B_t$, $E e^{zB_t}=e^{tz^2/2}$ for all $z\in \mathbb{C}$. The proof uses the fact that $E^{izB_t}=e^{-tz^2/2}$ and $e^{ay}e^{-y^2/2}$ is integrable on $\mathbb{R}$, and analytic functions agreeing on a real line must agree on the whole complex plane. However, I do not understand why $z \mapsto E(e^zB_t)$ is analytic in the first place. I would greatly appreciate any help.