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Suppose that $X$ is a random variable that follows a normal distribution $N(0,\sigma^2)$. We know that its characteristic function can be computed as follows $$ \mathbb{E}[e^{i t X}]=e^{-\frac{1}{2}t^2 \sigma^2}\quad\text{ for }t\in\mathbb{R}. $$ My question is: can this identity be extended to $t$ complex? that is, is it true that $$ \mathbb{E}[e^{i z X}]=e^{-\frac{1}{2}z^2 \sigma^2}\quad\text{ for }z\in\mathbb{C}? $$ Why?

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Yes. It is easy to show that $ t \to Ee^{itX}$ is well defined and differentiable on the entire complex plane. Since both sides of the equation you are trying to prove are entire functions which agree on the real line they automatically agree on the complex plane. [ $E|e^{itX}|\leq Ee^{|t||X|}<\infty$ for all complex numbers $t$. You can use DCT to prove differentiability. I will provide a detailed proof if necessary].

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