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I was just wondering if the rules of exponents still applied to imaginary and complex numbers, like if $(2^4)^i=2^{4i}$ or not and if $(4^i)^i=4^{-1}$, etc

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Short version: Yes, but you have to be careful. – Alex Becker Sep 15 '12 at 1:42
If you write the complex number in polar form $z=re^{i\theta}$ you can see that $(re^{i\theta})^i=re^{i^2\theta}=re^{-\theta}$. – axblount Sep 15 '12 at 1:43
Why is $r$ not raised to the $i$ in these comments? – alex.jordan Sep 15 '12 at 3:19

I think some of the comments are in error.

If $z = r e^{i \theta}$ with $r$ a positive real (if $z = 0$, then $z^i = 0$), $z = r e^{i (\theta + 2 \pi k)}$ for any integer $k$, so $z^i = r^i e^{i^2 (\theta + 2 \pi k)} = e^{i \ln r} e^{- (\theta + 2 \pi k)} = (\cos( \ln r) + i \sin(\ln r)) e^{- (\theta + 2 \pi k)} $.

The principal value is usually the one with $k = 0$, but all the other values are possible.

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The exact same rules apply. A nice example of this is a typical Oxbridge interview question;

$$ What\ is\ i^i\ ? $$

$$ i^i = {(e^{i\frac{\pi}{2}})} ^ i = e^{i^2\frac{\pi}{2}} = e^{-\frac{\pi}{2}} \approx 0.20788 $$

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I asked my math teachers this question when I was about 12. It took some time to get an answer. – Hew Wolff Sep 15 '12 at 2:39

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