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I've been reading up on why the value of 0^0 is controversial (see Zero to the zero power - is $0^0=1$?) and I wondered: is it possible for $i^i$ to have a value? I plugged it into a TI-83 calculator and it returned 0.2078795764 (!) How is this possible and why is the result a real number not a complex number?

Update I realize how to use Euler's Identity of $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ and I understand the oldest answer's value of $\theta = \pi/2$ to solve for $i^i$, but doesn't this mean I can pick any odd multiple of $\pi/2$ (such as $3\pi/2, 5\pi/2$, etc.) as a value of $\theta$? Does that mean that $i^i$ is somehow "periodic" like the cosine curve is?


marked as duplicate by tomasz, Old John, dfeuer, Daniel Fischer, ncmathsadist Oct 30 '13 at 23:32

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    $\begingroup$ This in a different question, with a different answer. $\endgroup$ – PyRulez Oct 30 '13 at 22:11
  • $\begingroup$ $i^i$ is, strictly speaking, whatever you want it to be. Just because there are definitions and properties for $R^R$ and $R^I$ does not mean that there is a definition for $I^I$ (for Real R and Complex I). The real question is "how is $I^I$ defined?" and "what laws of exponents is that definition consistent with?" $\endgroup$ – DanielV Oct 30 '13 at 22:59

$i=e^{i(\pi/2+2k\pi)}$ therefore $i^i=(e^{i(\pi/2+2k\pi)})^i=e^{-\pi/2-2k\pi}$

for $k=0$ you get $0.207879576350761908...$

for $k=1$ you get $0.000388203203926766...$



$i=1\times i$, then $i^i=1^i\times i^i$: as tomasz commented, this post may be helpful: What is the value of $1^i$?

  • $\begingroup$ Look up Euler's formula: en.m.wikipedia.org/wiki/Euler's_formula $\endgroup$ – frogeyedpeas Oct 30 '13 at 21:19
  • $\begingroup$ I forgot about Euler's Identity. Nice! $\endgroup$ – Xoque55 Oct 30 '13 at 21:26
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    $\begingroup$ By your reasoning, $i=e^{5i\pi/2}$, therefore $i^i=e^{-5\pi/2}=e^{-\pi/2}$, so $e^{-4\pi/2}=1$ and $-4\pi/2=0$. $\endgroup$ – tomasz Oct 30 '13 at 21:46
  • $\begingroup$ @tomasz probably there are many different values for $i^i$ $\endgroup$ – newzad Oct 30 '13 at 22:01
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    $\begingroup$ I am pretty sure they define it be when $k=0$. That makes it agree with the normal definition. (After all, we usually like $4^{\frac12}$ to have a definite value, right? $\endgroup$ – PyRulez Oct 30 '13 at 22:08

That's not the only answer in fact.

For any $\alpha\in\mathbb{C}$, we define, for any complex $z=a e^{i\theta }\neq 0$, the multifunction $$[z^{\alpha }]:=\{e^{\alpha (\log |z|+i\theta )} : \theta\in\{\arg (z)+2k\pi i: k\in\mathbb{Z}\}\},$$ taking $z^{\alpha }$ as $w=be^{i\phi }\in [z^{\alpha }]$ such that $\phi\in (-\pi , \pi ]$; we take the "principle value" in the set defined above. [Recall that $e^{2\pi }=1.$.]

Take $z=\alpha =i$.

[This definition is from "Introduction to Complex Analysis: second edition," by H. A. Priestley.]

  • $\begingroup$ I'm sorry about all the typos; it's been a long day and I'm using my phone. $\endgroup$ – Shaun Oct 30 '13 at 22:11

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