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Over the reals, I know this is true: $$(x^r)^s=(x^s)^r$$ for $x>0$ and $r,s\in\mathbb{Q}$.

Does the same equation hold over $\mathbb{C}$? I.e., $$(z^r)^s=(z^s)^r,$$ for $z\in\mathbb{C}-\{0\}$ and $r,s\in\mathbb{Q}$?

paul henri
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    Yes. But just as $y| y^2 = b$ may have two solutions $z^{n/m}$ may have $m$ possible values. and no need to set subtract the 0. This is true for all complex. – fleablood Sep 21 '16 at 08:07

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I assume by $z^{a/b}$ you mean the set of all $w \in \mathbb{C}$ satisfying $w^b = z^a,$ where $a$ and $b$ are assumed to be coprime integers. To have the question make sense, we also need to define $A^q = \{a^q : a\in A\}$.

Anyway, the answer is no. For instance: $$(1^{1/2})^2 = \{1,-1\}^2 = \{1^2,(-1)^2\} = \{1\}$$

$$(1^2)^{1/2} = 1^{1/2} = \{-1,1\}$$

goblin GONE
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