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.

  • 133
  • 3
  • 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/prove-that-ii-is-a-real-number) – S.C.B. Apr 07 '16 at 16:40
  • 2
    For 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 Answers3


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 $k'\in\mathbb{Z}$.

  • 8,650
  • 8
  • 22
  • 51

These are both multiple-valued expressions. Generally one computes $w^z$ as $e^{z\log w}$, so the multiple-valuedness 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 $$z_k=e^{i(\frac{\pi i}{2} -2k\pi i)}=\boxed{e^{\frac{\pi}{2}(4k-1)}} $$ for integral $k$.

Then, since the values of $\log z_k$ are $\frac{\pi}{2}(4k-1) - 2n\pi i=2\pi k -\frac{\pi}{2}-2n\pi i$, 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.

  • 40,421
  • 1
  • 29
  • 72

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: $$(i^i)^i = (e^{-\frac{\pi}{2}})^i = e^{-i\frac{\pi}{2}} = i^{-1} = i^{4-1} = i^3 = -i$$ I hope this helps.

  • 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