Does $(\omega^a)^b \neq \omega^{(ab)}$ for complex numbers $\omega, a, b \in \mathbb{C}$?

71 Views Asked by At

Im wondering wether the equality

$$(\omega^a)^b = \omega^{(a\cdot b)}$$

hold for general complex numbers $\omega, a, b \in \mathbb{C}$? I tried some specific values and the result seems fine, but when trying to prove the statement I receive double Logarithms/Exponentials and don't really know how to proceed..

1

There are 1 best solutions below

0
On

No, it does not hold in general.

COUNTEREXAMPLE: Consider $e^{i\theta}$ = $e^{i\theta 2\pi\over2\pi}$. If the stated identity were true (and symmetric), this would be equal to $(e^{2\pi i})^{\theta\over 2\pi}$ = $1^{\theta\over 2\pi}$, which would mean complex numbers do not exist.