Does $2^i$ exist, if so how do I calculate its value? 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? 

 A: 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.
A: We have: $2=e^{\log{2}}$
So, $2^i=\left( e^{\log{2}} \right) ^ i=e^{i \log{2}}=\cos(\log(2))+i \sin(\log(2))$.
A: $a^b=e^{b\ln a}$.  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.
