I've been struggling with this problem, actually I was doing a program in python and did

1j ** 1j(complex numbers) (In python a**b = $a^b$ )

and found out the answer to be a real number with value $0.2079$, How to calculate this value of $i^i$?

  • 2
    $\begingroup$ not really a duplicate of the first question-- this is a more specific question and the asker of the first seems to have less background. I also think it's not a duplicate of the second although the author would benefit from reading the second. $\endgroup$
    – hunter
    Nov 18, 2014 at 15:15
  • $\begingroup$ Why these would not be duplicates escapes me, I am afraid. The fascinating thing about these "$z^w$, $z$ and $w$ complex" questions is their repetition (hence, their is a demand) and the repetition of unsatisfying answers (from which one can deduce that the subject is horribly taught). $\endgroup$
    – Did
    Nov 18, 2014 at 19:32
  • 1
    $\begingroup$ @Integrator Right, let me teach then: There is no such thing as a function $z\mapsto z^w$ defined unambiguously (e.g. continuously, say) when $w$ is complex, not real integer. (Note that the Edit to your answer, posted after my comment above which may have caused the trouble, is odd since it alternates between the postmodern view that $z^w$ is a set and some old-fashioned identities where $z^w$ is indeed one complex number.) $\endgroup$
    – Did
    Nov 19, 2014 at 6:28
  • $\begingroup$ @Integrator Really? Then we can begin by this: what is $i^i$, according to you? A number, a set, a tiger, or a helicopter? $\endgroup$
    – Did
    Nov 19, 2014 at 13:33
  • $\begingroup$ @Integrator Maybe this is a language problem but I understand next to nothing to your comment. "I asked robjohn" You asked robjohn what? And robjohn said what? And why should robjohn be involved in this? "I don't really understand what that statement really means" My last comment mainly contains, not a statement but, a question: what is $i^i$? The question seems simple enough, what is there to "understand" in it? If you cannot answer it, this is another matter. $\endgroup$
    – Did
    Nov 19, 2014 at 14:41

2 Answers 2


First you need to realize that this is a multi-valued function.

$$i=0+i$$ $$i=\cos\left(\frac{(4k+1)\pi}{2}\right)+i\sin\left(\frac{(4k+1)\pi}{2}\right)$$ $k\in \mathbb Z$, Using Euler's formula $$e^{i\theta}=\cos\theta+i\sin\theta$$



$$\begin{align}i^i&=\left(e^{\large {i(4k+1)\pi}/{2}}\right)^i\\ &=e^{\large{i\times i(4k+1)\pi}/{2}}\\ &=e^{\large{-(4k+1)\pi}/{2}}\\ \end{align}$$

Depending on the branch cut (part of $\mathbb{C}$ excluded) and branch of log (what value $\log(z_0)$ has for some $z_0$ not in the branch cut), there is a $k\in\mathbb{Z}$ so that $i^i=e^{-(4k+1)\pi/2}$. Using the most standard branch cut, $i^i=e^{-\pi/2}$

And python didn't seem to care about that and returned

$$i^i=e^{\large{-\pi}/{2}}\approx 0.2078795$$

  • 7
    $\begingroup$ Remember that $i$ also equals $e^{5i\pi/2}$, which gives a different answer. $\endgroup$
    – Arthur
    Nov 18, 2014 at 14:50
  • $\begingroup$ @Arthur So how can we say that it is 0.2079 only $\endgroup$ Nov 18, 2014 at 14:51
  • $\begingroup$ Here's a related one on MO: mathoverflow.net/a/94833/13042 $\endgroup$ Nov 18, 2014 at 14:51
  • $\begingroup$ To expand on what @Arthur said, this depends on what the choice of argument you use. This is similar to the complex logarithm. $\endgroup$
    – Joel
    Nov 18, 2014 at 14:52
  • $\begingroup$ Considering Euler's formula $e^{i\pi} + 1 = 0$ I think Integrator's principal answer was correct. I left an answer, which describes how Euler's principal formula is used. $\endgroup$
    – Amad27
    Nov 18, 2014 at 14:58

$e^{i\pi} = -1$

$i = \sqrt{e^{i\pi}}$

$i^i = e^{-\frac{\pi}{2}}$

I just wanted to point out the derivation!

  • $\begingroup$ Why the downvote? $\endgroup$
    – Amad27
    Nov 18, 2014 at 14:59
  • $\begingroup$ But I used Euler's formula. $e^{i\pi} + 1 = 0$ then I derived the rest. $\endgroup$
    – Amad27
    Nov 18, 2014 at 15:05
  • $\begingroup$ This is not that bad, but I can equally ask, since $e^{i3\pi} = -1$, can you come up with a more general formula for $i^i$, or complex power in general? $\endgroup$
    – peterwhy
    Nov 18, 2014 at 15:19

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