Let $(X_n)_{n\ge 0}$ be a series of real random variables and $\phi_n(\lambda)=\mathbb E(e^{\lambda X_n})$, where $\lambda$ is a complex number. Consider the following two types of convergence:
(1) There exists an open neighborhood $U$ of $0$ on the real line such that $\phi_n(\lambda)$ converges to $\phi_0(\lambda)$ uniformly for $\lambda \in U$;
(2) There exists an open neighborhood $U$ of $0$ on the complex plane such that $\phi_n(\lambda)$ converges to $\phi_0(\lambda)$ uniformly for $\lambda \in U$;
(3) There exists an open strip-shaped neighborhood $U=I\times \mathbb R$ of $0$ on the complex plane such that $\phi_n(\lambda)$ converges to $\phi_0(\lambda)$ uniformly for $\lambda \in U$;
Clearly $(3)\Rightarrow (2) \Rightarrow (1)$. Are they actually equivalent (the first implies the third)?
I believe this one might be solved using Cauchy-Schwartz. Indeed, you have, for $\lambda = a + ib$, $a, b \in \mathbb{R}$,
$$ \mathbb{E}[e^{(a + ib) X_n}] \leq \mathbb{E}[e^{2a X_n}]^{1/2} $$ where I used that $|e^{i2bX_n}| = 1$ for the second factor of Cauchy-Schwartz. Hence if $a$ is small enough, you can use (1) to conclude that (3) holds.