Is $(-2)^{\sqrt{2}}$ a real number? 
Is $(-2)^{\sqrt{2}}$ a real number?

Clarification: Is there some reason why $(-2)^{\sqrt{2}}$ is not a real number because it doesn't make sense why it shouldn't be a real number.
Mathematically we can define the value of $\sqrt{2}$ in terms of a limit of rationals. But the problem is is that some sequences will have values that are not defined for $(-2)^q$ where $q$ is a rational, such as $\dfrac{3}{2}$. Is this the reason why we can't define it or is there some other reason?

$$\left(-2\right)^{\sqrt{2}}=\left(\left|-2\right|e^{\arg(-2)i}\right)^{\sqrt{2}}=\left(2e^{\pi i}\right)^{\sqrt{2}}=2^{\sqrt{2}}e^{\pi\sqrt{2}i}=2^{\sqrt{2}}e^{\left(2\pi k+\pi\sqrt{2}\right)i}$$
With $k\in\mathbb{Z}$
