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For complex numbers $a,b,c$, explain why $a^{b\cdot c}=(a^b)^c$ is not necessarily true.

I know that complex powers are really sets of complex numbers. But coming from real analysis the above seems confusing. Can anyone give a simple explanation of what is going on.

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    $\begingroup$ Everything is simple if you consider $a^b=e^{b \ln a}$. And remember that the $ln$ is multibranch function. $\endgroup$ Dec 4, 2017 at 6:19
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    $\begingroup$ Why do you expect it to be true? When $b$, $c\in{\mathbb N}$ it can be proven by induction from the intended meaning of $a^b$, but otherwise the definition of $a^b$ is in a way "artificial". $\endgroup$ Dec 4, 2017 at 10:46

2 Answers 2

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I'll provide a counterexample. Let $c=1/2$ and $b=2$. Now let $a=i$. $a^{bc}=i^{2*1/2}=i^{1}=i$. But, $a^{b}=-1$ and $(a^{b})^{c}=(-1)^{1/2}=-i \textrm{ or } i$. We see that raising a number to the one half is multivalued, so we have an issue here. So in this case, $a^{bc} \neq (a^b)^c$. The explanation for how to get around this has to do with Branch cuts (https://en.wikipedia.org/wiki/Branch_point#Branch_cuts}).

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EG

in $\mathbb{C}$ consider a=1, b=2 and $c=\frac12$:

$$a^{(bc)}=1^1=1$$

$$(a^b)^c=1^{\frac12}=\pm 1$$

conversely in $\mathbb{R}$ $$1^{\frac12}=+ 1$$

EG 2

consider a=i, b=2 and $c=\frac12$ then:

$$a^{bc}=i^1=i$$

$$(a^b)^c=i^{\frac12}=\pm e^{i\frac{\pi}{4}}$$

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