$(-1)^{\sqrt{2}} = ? $ This popped up when I was thinking about 
$$(-1)^{\frac {p}{q}}$$
where $ p $ and $q$ are integers such that 
 $\gcd (p,q) = 1$
If  $p$ is even :
$(-1)^{\frac {p}{q}} = +1$
If $q$ is even :
$(-1)^{\frac {p}{q}}$ = a complex power with base $i $
If both $p$ and $q$ are odd :
$(-1)^{\frac {p}{q}} = -1$
$\sqrt2 $ cannot be written in the form $\frac {p}{q} $. So I was wondering whether there is a way to find the answer. My guess is that we will never know because we will never get to know the complete decimal expansion of $\sqrt2$.
 A: Using Euler's formula:
$-1=\exp(\ln(-1))=\exp(\ln(1)+i\arg(-1))=\exp(0+i(2k+1)\pi) =\exp(i(2k+1)\pi)$
for $k\in \Bbb{Z}$
Thus:
$(-1)^{\sqrt{2}}=\exp(i(2k+1)\pi\sqrt{2})=\cos((2k+1)\sqrt{2}\pi)+i\sin((2k+1)\sqrt{2}\pi)$
A: You can write it as $$e^{\sqrt{2}\ln(i^2)}$$ now $\ln(i^2)=\ln(1)+i\arg(i^2)=0+i(2k+1)\pi$ thats it. You can continue now. Log of complex number is $log(z)=log(|z|)+arg(z)$
A: If you draw the unit circle, labeling the y-axis as complex and the x-axis as real, then $(-1)^{\sqrt2}$ has an angle of $\theta=180\sqrt2\approx254$ degrees and absolute value of $1$,

not drawn to scale
If you know trigonometry, the Cartesian coordinate of that spot is given as $(-1)^{\sqrt2}=\cos(254)+i\sin(254)$.
If you are interested in the actual reasoning behind this result, I would recommend looking up DeMoivre's formula, or proving DeMoivre's formula by calculating $(1+i)^n$ for $n=1,2,3,\dots$ and writing the distance from $0$ and the angle measures.  Then apply what you discovered about DeMoivre's formula on your problem.
