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I feel like there is a pretty critical flaw with my logic here but I cant figure out where it is.

I don't have much knowledge about imaginary and complex numbers , but I've watched Khan academy's video on them earlier and saw this video here , where he demonstrated that very high powers of $i$ can be simplified by taking multiples of 4 of them , since $i^4=1$

say we have $i^{25}$, cant we rewrite this as $i^{4*\frac{25}{4}} = (i^4)^{\frac{25}{4}} = 1^{\frac{25}{4}} = 1$ ? and do this for every number that belongs to R?

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    $\begingroup$ $z\mapsto z^4$ is not injective over $\mathbb{C}$, hence for any $z\in\mathbb{C}^*$ there are four distinct complex numbers $w$ such that $w^4=z$. You have to be careful in writing $z^{1/4}$. $\endgroup$ – Jack D'Aurizio Nov 1 '17 at 20:41
  • $\begingroup$ so basically it isn't as simple as I'm making it out to be? $\endgroup$ – Dahen Nov 1 '17 at 20:43
  • $\begingroup$ $(-1)^{2*\frac{25}{2}}=((-1)^2)^{\frac{25}{2}}=1^{\frac{25}{2}}=1$? $\endgroup$ – Eclipse Sun Nov 1 '17 at 20:46
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    $\begingroup$ The "obvious" identity $(z^a)^b=z^{ab}$ works for $z\in\mathbb{R}^+$, but not in general. $\endgroup$ – Barry Cipra Nov 1 '17 at 20:52
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    $\begingroup$ @Dahen: the issue relies in defining $z^{1/4}$ or $z^{1/3}$ when $z$ is not a non-negative real number, since you have to deal with the lack of injectivity of some function. $\endgroup$ – Jack D'Aurizio Nov 1 '17 at 20:53

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