If $$ e^{ix} = \cos{x} + i \sin{x} $$ which means $$ e^{2\pi i} = \cos{2\pi} + i \sin{2\pi} = 1 + 0i = 1 $$ and $$ a^{(bc)} = (a^b)^c $$ then why is this wrong for some real number $x$? $$ e^{ix} = e^{2\pi i(x/2\pi)} \\ = (e^{2\pi i})^{(x/2\pi)} \\ = 1^{(x/2\pi)} \\ = 1 $$ ?
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Cf. this question: fractional powers of negative numbers are not uniquely defined, and the "general rule" $(a^b)^c=a^{b×c}$ does not always work when $b$ and $c$ are not integers – J. W. Tanner Dec 25 '20 at 20:03
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The "rule" $a^{bc}=(a^b)^c$ simply does not hold for complex numbers. Indeed, it doesn't even hold for all real numbers (consider $a=-1$ and $b=2$ and $c=\frac12$).
This is a great example of how, as mathematicians, we need to remember more than just the equations we learn—we need to remember the precise conditions under which the equations hold.
Greg Martin
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So when can I count on the rule being true? I assume if $a$ is a positive real and $b$ and $c$ are real. Also if $b$ and $c$ are both integers, right? When else? What if one of $b$ or $c$ is an integer? – ajb Dec 26 '20 at 04:27