# Complex numbers on the values of i

I am trying to calculate:

1. $i^2 \div 3$
2. $i^i$
3. $(i+1)^{i-1}$.

For the first one, $i = \frac{1}{2}\cdot e^i\cdot\pi$.

So, $(\frac{1}{2}\cdot e^i\cdot\pi)^2 \div 3$, and I tried this trick for the other two but it is not getting me anywhere.

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At least i^i was the subject of a recent question (which got the proper answer one can imagine) and maybe more of them. – Did Sep 15 '12 at 22:19
I hope you mean $i = e^{i\pi/2}$. And was (1) supposed to be $i^{2/3}$? – Robert Israel Sep 16 '12 at 5:45
for (1) I wanted i^2/3 and can you do the explanation without using cis? Thanks – mary Sep 16 '12 at 6:19
You really wanted $i^2/3$, that is, $(i^2)/3$? Do you know what $i^2$ is? – Robert Israel Sep 16 '12 at 6:38

The general definition of $a^b$ (for $a \ne 0$) is $e^{b \log a}$. But you have to be careful with this because it is a multivalued function if $b$ is not an integer: $\log a$ has infinitely many values. So for example $\log(i) = i (\pi/2 + 2 n \pi)$ for arbitrary integers $n$. Thus in (2), $$i^i = e^{i \log(i)} = e^{-(\pi/2 + 2 n \pi)}$$