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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..

user7802048
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  • The short answer is that complex exponentials are defined as $\omega^a = \exp (a \log \omega)$, defining this uniquely requires an arbitrary choice of the branch of $\log$ that may not "work" for all values of the exponent. – Connor Harris Sep 23 '19 at 14:56

1 Answers1

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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.

Certainly not a dog
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