Is $(-3)^\sqrt{2}$ a real number ? If it a complex number, then what is its $a+bi$ form? Is this a real number? If it a complex number, then what is its $a+bi$ form?
$$(-3)^\sqrt{2}$$
 A: As a complex number, it has infinitely many values.  We have $$(-3)^\sqrt{2}=e^{\sqrt{2}\log(-3)}=e^{\sqrt{2}(\ln{3}+(2n+1)\pi i)}=3^{\sqrt{2}}e^{\sqrt{2}(2n+1)\pi i}\\=3^{\sqrt{2}}\cos(\sqrt{2}(2n+1)\pi)+i3^{\sqrt{2}}\sin(\sqrt{2}(2n+1)\pi),\\n=0,\pm1,\pm2,\dots$$
A: Exponentiation of a nonzero complex base $b$ with any rational number $r$ is always well-defined by letting $b^{p/q} = \frac{b^p}{b^q}$. For positive bases $b > 0$, we can extend the domain of $b^x$ for all real $x$ in two equivalent ways. The first is to notice that $b^x$ is continuous on $\mathbb{Q}$, and since the rationals are dense in $\mathbb{R}$ there is a unique way to extend the function to a continuous function on all of $\mathbb{R}$. The second is to fix the natural exponentiation function $e^x$ for any real-valued $x$, and then let $b^{x} = e^{x\ln{b}}$.
For a negative base $b$ (or more generally any non-positive complex number), neither of these definitions work since $b^x$ is no longer a continuous function on the rationals, and $\ln(x)$ is not defined. However, using the $complex$ logarithm $Log$, we can still use the second method to assign a value to $b^x$ for any complex numbers $b\neq 0$ and $x$ by again letting $b^{x} = e^{x\cdot Log{b}}$. However, the issue here is that the complex logarithm is not actually a function since it is multi-valued; for any number $z$ such that $e^z = w$, then $e^{z+2\pi i n} = w$ for any integer $n$ and so $z + 2\pi i n$ are all `logarithms' of $w$. Therefore, while we can assign a value to $(-3)^{\sqrt{2}}$, it is not unique; there are infinitely many such values we could have chosen. 
Just to illustrate, notice that $e^{\ln{3} + \pi i} = 3e^{i\pi} = -3$ and so $\ln{3} + \pi i$ is a complex logarithm for $-3$. Thus, fixing this choice of logarithm, $$(-3)^{\sqrt{2}} = e^{\sqrt{2}\cdot (\ln{3} + \pi i)} = 3^{\sqrt{2}}e^{\pi\sqrt{2}i} = 3^{\sqrt{2}}(\cos(\pi\sqrt{2}) + i\sin(\pi\sqrt{2}))$$. However, we could have just as easily calculated that $$(-3)^{\sqrt{2}} = 3^{\sqrt{2}}(\cos(\pi(1+2n)\sqrt{2}) + i\sin(\pi(1+2n)\sqrt{2}))$$ for any integer $n$.
