Is $e^{(2i\ln b)}=(e^{\ln b})^{2i}=b^{2i}$ where $b$ is a real positive number?

Question

Here, $b$ is a positive real number. Does the equalities below holds true? The complex exponential is quite messy.

$$e^{(2i\ln b)}=(e^{\ln b})^{2i}=b^{2i}$$

I believe there is some multiple of $2i\pi$ missing somewhere.

Answer 1

This is totally fine. There is nothing missing.

I think you meant something like

$\ln(-a) = (2n+1)*\pi*i +\ln(a)$

for every integer n.