This is a silly question but under what conditions is $a^{xy}=(a^x)^y$ true, given all are complex numbers?
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If $a\ne 0$. If $a=0$, $x>0$, $y>0$ – Michael Galuza Jul 16 '15 at 14:52
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Uh... it's an identity, it's always true unless something is not well-defined. – msinghal Jul 16 '15 at 14:53
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@Michael Galuza, What if a=-1, x=2, y=.5? – Sameer Kulkarni Jul 16 '15 at 15:06
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Let's spell out the definitions. For complex numbers $z$ and $\alpha$, and choice of range of argument, such as $[0,2\pi)$ or $(-\pi, \pi]$, we can define $$z^{\alpha} = \exp(\alpha \log z) = \exp(\alpha (\log |z| + i\arg z))$$ With this definition, we have $$a^{xy} = \exp(xy(\log |a| + i\arg a))$$ Now $a^x = \exp(x(\log |a| + i\arg a))$, so $\log |a^x| = \log (\exp(x \log |a|)) = x\log |a|$ and $\arg(a^x) = \arg a + 2n\pi$ for the unique $n_x \in \mathbb{Z}$ making $\arg(a^x)$ live in the desired range. Hence $$(a^x)^y = \exp(xy(\log |a| + i(\arg a + 2n_x\pi))$$
In other words, $$(a^x)^y = a^{xy}\exp(2n_xxyi\pi)$$ for some $n \in \mathbb{Z}$. Thus we have $(a^x)^y = a^{xy}$ if and only if $n_x=0$ or $xy \mid n_x$.
If you consider complex exponentiation as a multifunction, the identity $(a^x)^y = a^{xy}$ is true since this issue with the choice of range of argument goes away.
Clive Newstead
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I've noticed a few typos in this since I posted it, but the idea is there; I also think I rely on $x \in \mathbb{R}$ here; for general $x \in \mathbb{C}$ I think it becomes even more complicated! If I remember, I'll fix it later. – Clive Newstead Jul 16 '15 at 15:24