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I don't understand what is wrong with the following reasoning. Can someone help me understand? $f:\mathbb{R}\to \mathbb{C}$ by $f(t) = e^{2\pi ~i ~t} = ((e^{i\pi})^2)^{t} = ((-1)^2)^{t} = 1^{t} = 1$

I don't see where the computation goes wrong, but it is obviously wrong somewhere.

Blue
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Jim Newton
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1 Answers1

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$x^{yz}=(x^y)^z$ is only valid for $x\in\mathbb{R}_+$ and $y,z\in\mathbb{R}$.

For further information, have a look at Question on roots of unity .

Andreas Tsevas
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    I would like to know under which conditions does $x^{yz} = (x^y)^z$. I don't think that is answered in https://math.stackexchange.com/questions/437748/question-on-roots-of-unity, at least not explicitly. There was a comment above that this is only true if x is real. However, it is sometimes not even true when x is real. – Jim Newton Jul 19 '22 at 11:55
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    An alternative problem is that the complex exponential (and exponentials in general) only behave how we want them to (i.e. allowing breaking exponents into products) if we treat them as multivalued maps, and when viewed this way the problem is that $1^t$ is not identically $1$ unless $t$ in an integer. – Fishbane Jul 19 '22 at 11:56
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    @JimNewton One difficulty in answering that question is that it ultimately depends on how you define exponentials. However the only case where it is true independent of definition is if both $y$ and $z$ are integers then $x^{yz}=(x^y)^z$. – Fishbane Jul 19 '22 at 12:04
  • @Fishbane, as this post is marked a duplicate, I've created another question to address this issue (as suggested by the stackexchange duplication mechanism). https://math.stackexchange.com/questions/4495918/when-does-xyz-xyz – Jim Newton Jul 19 '22 at 12:11