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I'm a high school senior and this question has been on my mind since 8th grade, since I've learned Euler's formula but I've thought about it a lot more recently. According to Wolfram Alpha: (e^i)^(2iπ)=e^(-2π)

(e^(2iπ))^(i)=1

How is this possible??????

In other words, if

$$ e^{iπ} = -1 $$

$$ e^{2iπ} = 1 $$

$$ (e^{2iπ})^i = 1^i $$

$$ e^{-2π} = 1 $$

Where is the error in this reasoning?

Physor
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neuraljuror
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    Many of the identities you think you know about exponents should have come with a warning that they only are guaranteed to work when using non-negative real bases and real exponents. $(a^b)^c$ is not necessarily the same as $a^{bc}$ – JMoravitz Oct 09 '20 at 13:36
  • What is your definition of $z^w$ for complex $z$ and $w$? And did you actually check that exponentiation of complex numbers satisfies all the algebraic rules of exponentiation of positive real numbers? – KCd Oct 09 '20 at 13:40
  • See also here – JMoravitz Oct 09 '20 at 13:41
  • Please use MathJax to format your equations – Andrei Oct 09 '20 at 13:41
  • @JMoravitz Thanks so much! Yes, that definitely answered my question and the article was really helpful. – neuraljuror Oct 09 '20 at 13:46

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