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I was working on the question, to show that $i^i$ is a real number.

That was however straight forward,

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

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

So I thought I'd similarly work out the value of $2^i$ or $3^i$, but got stuck,

$$2^i = (2e^{i0})^{i}$$ $$= 2^i e^{0}=2^i $$

which does not help, so I tired,

$$2^i = (2e^{2\pi i})^{i}= 2^ie^{-2\pi }$$

which is utter nonsense (and bad maths?)

So google gives a value of, $$2^i = 0.769238901 + 0.638961276 i$$

So my question is where does that come from? How can one compute a complex exponent?

IRSAgent
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3 Answers3

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One generally computes $z^w$ as $e^{w\log z}$, which typically has a countably infinite set of values due to the $2\pi i$-periodicity of the exponential function. In other words, the logarithm is only defined up to an integral multiple of $2\pi i$.

So the possible values of $i^i$ are really $$e^{i(\pi i/2- 2\pi i k)}=e^{(4k-1)\pi/2}$$

for integral $k $. Likewise, the possible values of $2^i$ are $$e^{i(\ln 2- 2\pi i k)}=e^{2\pi k + i\ln 2}=e^{2\pi k}\cos\ln 2 + ie^{2\pi k}\sin\ln 2$$ for integral $k$ (here "$\ln$" denotes the ordinary real-valued natural logarithm of a positive real number).

All of the other answers so far are incomplete in that they omit all but one of a countably infinite number of values.

MPW
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    This is the only rigorous answer indeed. One has to be careful with complex exponentiation since e.g. $1=e^{i2\pi}$ does not imply the series of equalities $1=1^i=\left(e^{i2\pi}\right)^i=e^{-2\pi}$, which is the root of the error in the OP. – Ian Sep 28 '14 at 00:37
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    "All of the other answers so far are incorrect in that they omit all but one of a countably infinite number of values. " No, you're only right in the case of a non-positive or non-real base. In the case where the base is a real positive number, it's pretty clear cut what the answer is and there is no ambiguity. ( http://en.wikipedia.org/wiki/Exponentiation#Complex_exponents_with_positive_real_bases_2 ) Example: By your logic, $\log(e)=1+2 i \pi k$, so $e^{ix}=e^{2\pi k}\cos(x) + ie^{2\pi k}\sin(x)$. – Fujoyaki Sep 28 '14 at 00:59
  • $1^i = e^{i\log(1)}=e^{i(\ln(1)+2\pi i\mathbb{Z})} = e^{-2\pi \mathbb{Z}}$. This is a set of values which forms a sequence of points on $\mathbb{R}$. In short, @Fujoyaki it is quite clear that this positive real base gives an infinite set of values. This is standard stuff. See any complex analysis text. – James S. Cook Sep 28 '14 at 02:29
  • @JamesS.Cook I know about complex analysis, and I know about the contradiction that arise when you try exponentiation with an arbitrary complex base and an arbitrary complex exponent, no matter what branch of the logarithm you pick. However, for real positive bases (say $x$), complex exponentiation is defined by $x^z=e^{z \log x}$ where $\log x$ is real. In this special case, the definition bears no contradiction. ( http://en.wikipedia.org/wiki/Exponentiation#Complex_exponents_with_positive_real_bases_2 ) – Fujoyaki Sep 28 '14 at 02:59
  • @Fujoyaki granted. But, there are many additional solutions we ought to mention since the OP seems not to be aware of the difficulties of the bigger picture of complex powers. – James S. Cook Sep 28 '14 at 03:16
  • @JamesS.Cook Yeah, sure, I agree too, because he doesn't seem to know what he's doing when computing $i^i$, so I liked this answer for clarifying things up to the OP. But I still didn't like the fact that he dismissed all other answers as wrong, because they aren't. – Fujoyaki Sep 28 '14 at 03:20
  • @Fujoyaki: No offense intended. I have softened my statement to "incomplete" rather than "incorrect". – MPW Sep 28 '14 at 13:38
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We have: $2=e^{\log{2}}$

So, $2^i=\left( e^{\log{2}} \right) ^ i=e^{i \log{2}}=\cos(\log(2))+i \sin(\log(2))$.

James S. Cook
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Fujoyaki
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$a^b=e^{b\ln a}$. And $e^{c+id}=e^c\times e^{id}=e^c\times(\cos d+i\sin d)$ (this is Euler's Formula). So this gives: $$2^i=e^{i\ln2}=e^0\times(\cos(\ln 2)+i\sin(\ln2))=\cos(\ln 2)+i\sin(\ln2)$$ and that gets you the nasty numbers Google got you.

Dan Uznanski
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