What is $i^i$ and $(i^i)^i$. I fond some answers online but is there a clearer explanation of this? Maybe something help me to understand.

You are not familiar with Euler's identity? – Alex S Apr 07 '16 at 16:38

No, could you recommend me some text for this? Thanks. – JOHN Apr 07 '16 at 16:38

Using the representation that $i = e^{i \pi/2}$, we have $i^i = \left(e^{i\pi/2}\right)^i = e^{i^2\pi/2} = e^{\pi/2}$. Your problem seems to be answered [here](http://math.stackexchange.com/questions/191572/provethatiiisarealnumber) – S.C.B. Apr 07 '16 at 16:40

2For complex numbers $z$, we define $a^z := \exp (z \log a)$, but in general this depends on a choice of branch cut of the logarithm function. – Travis Willse Apr 07 '16 at 16:49

This has already been discussed to death several times on the site... – Did Apr 07 '16 at 17:05
3 Answers
In general, complex power of $a^b$ $(a\ne 0)$ is defined as $$ a^b = e^{b\log a}. $$ Thus $$ i^i = e^{i\log i}=e^{i(\operatorname{Ln}1+i(\frac{\pi}{2}+2k\pi))}=e^{\frac{\pi}{2} 2k\pi} $$ for $k\in\mathbb{Z}$ and $$ (i^i)^i =\left(e^{\frac{\pi}{2} 2k\pi}\right)^i=e^{i\log\left(e^{\frac{\pi}{2} 2k\pi}\right)}=e^{i(\frac{\pi}{2}2k\pi+i2k'\pi)}=ie^{2k'\pi} $$ for $k'\in\mathbb{Z}$.
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These are both multiplevalued expressions. Generally one computes $w^z$ as $e^{z\log w}$, so the multiplevaluedness comes from that of the logarithm.
First, since the values of $\log i$ are $\frac{\pi i}{2} 2k\pi i$, the values of $i^i$ are $$z_k=e^{i(\frac{\pi i}{2} 2k\pi i)}=\boxed{e^{\frac{\pi}{2}(4k1)}} $$ for integral $k$.
Then, since the values of $\log z_k$ are $\frac{\pi}{2}(4k1)  2n\pi i=2\pi k \frac{\pi}{2}2n\pi i$, the values of each $z_k^i$ are $$z_{k,n}=e^{i(2\pi k \frac{\pi}{2}2n\pi i)} =e^{2n\pi}\cdot e^{2k\pi i \frac{\pi i}{2} } = ie^{2n\pi} $$ for integral $n$ and $k$. As it happens, for a given $n$, the values of $z_{n,k}$ coincide for each integral $k$, so we may suppress the unneeded subscript. Then we have the possible values of $(i^i)^i$ are $$\hat{z}_{k} = \boxed{ie^{2k\pi}}$$ for integral $k$.
Note that the previous answers are incomplete.
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Start by representing $i$ in exponential form via the Euler relation: $$i = e^{i\frac{\pi}{2}}$$ Then combine exponents to get the answer: $$i^i = e^{i * i\frac{\pi}{2}} = e^{\frac{\pi}{2}}$$ To get to your second term, combine exponents and then take advantage of the fact that powers of $i$ form a group with integer exponents modulo 4: $$(i^i)^i = (e^{\frac{\pi}{2}})^i = e^{i\frac{\pi}{2}} = i^{1} = i^{41} = i^3 = i$$ I hope this helps.
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This is not correct, because $z^{\alpha\beta}=(z^{\alpha})^{\beta}$ does not hold, in general. – choco_addicted Apr 07 '16 at 17:12

@choco_addicted: More properly, these are both incomplete. The state values are indeed valid values of these expressions, but there are infinitely many distinct values of both. – MPW Apr 07 '16 at 17:30