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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 $i^i$?

Aditya Hase
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Pruthvi Raj
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    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. – hunter Nov 18 '14 at 15:15
  • 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). – Did Nov 18 '14 at 19:32
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    @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.) – Did Nov 19 '14 at 06:28
  • @Integrator Really? Then we can begin by this: what is $i^i$, according to you? A number, a set, a tiger, or a helicopter? – Did Nov 19 '14 at 13:33
  • @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. – Did Nov 19 '14 at 14:41
  • ((Three comments by @Integrator now deleted. I don't know what is happening here.)) – Did Nov 19 '14 at 14:43
  • @Integrator Sorry but, if you delete your comments once I answered them, where are we going? All this begins to look awfully like an evasion tactics (and I am not much interested in this game). – Did Nov 19 '14 at 14:45
  • @Integrator No. – Did Nov 19 '14 at 14:47
  • ((Two other comments by @Integrator now deleted.)) – Did Nov 19 '14 at 18:37

2 Answers2

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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 $$e^{i\theta}=\cos\theta+i\sin\theta$$

$$i=e^{\large{i(4k+1)\pi}/{2}}$$

Now

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

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

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

Aditya Hase
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    Remember that $i$ also equals $e^{5i\pi/2}$, which gives a different answer. – Arthur Nov 18 '14 at 14:50
  • @Arthur So how can we say that it is 0.2079 only – Pruthvi Raj Nov 18 '14 at 14:51
  • Here's a related one on MO: http://mathoverflow.net/a/94833/13042 – Giuseppe Negro Nov 18 '14 at 14:51
  • To expand on what @Arthur said, this depends on what the choice of argument you use. This is similar to the complex logarithm. – Joel Nov 18 '14 at 14:52
  • 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. – Amad27 Nov 18 '14 at 14:58
  • @Integrator So is this the answer or shall I have to go through the choice of argument thing to find the correct answer? – Pruthvi Raj Nov 18 '14 at 15:01
  • @Integrator, deserves downvotes for this new answer. Most questions asked depend on the interval of principal interval. For sine, it is $[-2\pi, 2\pi]$, similar to that. – Amad27 Nov 18 '14 at 15:01
  • @PruthviRaj, I dont think this is correct. Most answers including wolframalpha take the principal values. – Amad27 Nov 18 '14 at 15:02
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$e^{i\pi} = -1$

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

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

I just wanted to point out the derivation!

Bman72
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Amad27
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