I know that numbers with rational power can be converted to radicals and based on the degree of the radical we can say that whether they are real or complex.
But what about numbers like $\left ( -2 \right )^{\sqrt{2}}$ or $\left ( -2 \right )^{\pi}$ that have irrational powers , what can we say about them ?
I tried Google calculator and it gave me these but I don't know why :
$$\left ( -2 \right )^{\pi}=-7.9661783 - 3.7973987 i$$
$$\left ( -2 \right )^{\sqrt{2}}=-0.709608865 - 2.56893919 i$$
On
The number you specified is not at all a number. Following the standard definitions you have that $a^b=e^{b\ln a}$. In your case, since $\ln (-2)$ is not defined (when thinking of $\ln$ is a function of real numbers), the exponent does not exist. Now, in complex analysis $\ln$ be be extended to non-zero complex numbers but the extension is not unique. Thus $\ln (-2)=w$ if, and only if, $e^w=-2$ and this equation has infinitely many different solutions in the complex plane. So, $(-2)^{\sqrt 2}$ is actually a whole set of numbers, none of which is a real numbers. If you consult any elementary text on complex numbers it would treat complex exponentiation.
If $x$ is a negative real number, then $x^y$ can be evaluated as a real number in general only if $y$ is a rational number with an odd denominator.
I say "can be" because, as Ittay notes in his answer, $x^y$ is actually a multivalued function, in general. The 'principal' value for $(-1)^{1/3}$ is actually not real, but $(-1)^{1/3}$ can be evaluated as real because there is a real value in the set of values.
If $x$ is a negative real number, then the values of $\log x$ can be an of the values $\log(-x) + i(2k+1)\pi$ for some integer $\pi$. Then you get the possible values:
$$x^y = e^{y\log x} = e^{y\log(-x) + iy(2k+1)\pi} = (-x)^y \left(\cos y(2k+1)\pi + i\sin y(2k+1)\pi\right)$$
This is real if and only if $y(2k+1)\pi$ is an integer multiple of $\pi$. We can find such a $k$ if and only if $y$ is a rational number with an odd denominator.