$\Bbb{E}(e^{\lambda X})=e^{\lambda^2/2}$ impliesthat $\Bbb{E}(e^{zX})$ is holomorphic and well-defined. I am reading Brownian Motion, Martingales, and Stochastic Calculus. The author starts with Gaussian Random Variables.
He say that 

The complex Laplace transform of X is then given by $\Bbb{E}(e^{zX})=e^{z^2/2}$ for all $z\in \Bbb{C.}$ To get this formula (and also to verify that the complex Laplace transform is well
defined), consider first the case when $z=\lambda\in \Bbb{R}$: $$\Bbb{E}(e^{\lambda X})=\ldots=e^{\lambda^2/2}$$

This calculation ensures that $\Bbb{E}(e^{zX})$  is well-defined for  $z\in \Bbb{C}$, and defines a holomorphic function on $\Bbb{C}$.


Question:
Does it means that if we consider real and
imaginary parts separately then it follow that $\Bbb{E}(e^{zX})$  is well-defined and as the result for both real and imaginary parts is a holomorphic function it follow that $\Bbb{E}(e^{zX})$  is holomorphic ?
 A: A short answer. Both real and imaginary parts of the m.g.f. are well-defined. But they are not holomorphic separately.
A longer answer. Let $X$ be a $\mathbb{R}$-valued random variable such that its m.g.f. $M_X(s) = \mathbb{E}[e^{sX}]$ converges for $s \in I$ for some open interval $I$ containing $0$. Then utilizing the fact that $|e^z| = e^{\operatorname{Re}(z)}$, on the strip $I \times \mathbb{R}$ we have
$$\forall z = s + i\xi \in I \times \mathbb{R} \ : \qquad
\mathbb{E}\left[\left|e^{zX}\right|\right]
= \mathbb{E}\left[e^{sX}\right]
= M_X(s)
< \infty$$
and thus $M_X$ is well-defined on $I \times \mathbb{R}$. Next, for any closed piecewise $C^1$ curve $\gamma : [0, 1] \to I \times \mathbb{R}$,
\begin{align*}
\int_{0}^{1} \mathbb{E}\left[ \left| e^{\gamma(t) X} \gamma'(t) \right| \right] \, dt
&\leq \int_{0}^{1} \mathbb{E}\left[ e^{\operatorname{Re}(\gamma(t)) X} |\gamma'(t)| \right] \, dt \\
&\leq \operatorname{length}(\gamma) \cdot \sup_{z \in \gamma}M_X(\operatorname{Re}(z)) < \infty.
\end{align*}
This allows us to apply the Fubini's theorem to obtain
$$ \int_{\gamma} M_X(z) \, dz
= \mathbb{E}\left[  \int_{\gamma} e^{zX} \, dz \right]
= 0. $$
So by the Morera's theorem, $M_X$ is holomorphic on $I \times \mathbb{R}$.
