How to compute $E[e^{(i-1)Z}]$ where $Z$ is standard normal

Question

Is there a trick on how to compute $$E[e^{(i-1)Z}]$$ where $Z$ is standard normal and $i$ is the imaginary number. We know that $E[e^{iZ}]$ is just a characteristic function of $Z$ but this is a little bit different.

Answer 1

You can calculate $E[\exp(zZ)]$ for any arbitrary complex number $z$. Note that all odd moments of $Z$ are $0$, while the even moments are given by $E[Z^{2k}] = \frac{(2k)!}{2^k k!}$. The power series of the exponential function and the dominated convergence theorem yield the following equality: $$E[\exp(zZ)] = \sum \limits_{l = 0}^\infty \frac{1}{l!} z^l E[Z^l] = \sum \limits_{k = 0}^\infty \frac{1}{(2l)!}z^{2k} E[Z^{2k}] = \sum \limits_{k = 0}^\infty \left(\frac{z^2}{2}\right)^k \frac{1}{k!} = \exp\left(\frac{z^2}{2}\right)$$