What is $(-1)^\sqrt{2}$?

Question

How do you reason that $(-1)^\sqrt{2}$ is a complex number based on the following statements:

1. $(-1)^{p/q}$ is a complex number if $p$ is odd and $q$ is even integers.
2. Rational numbers are dense in the real line.
3. Exponential is a continuous function.

Another question arises as what $(-1)^\sqrt{2}$ represents on the real line?

Answer 1

Note that in general $z^c$ is defined as $e^{c\log(z)}$, where $\log(z)$ is the multi-valued complex logarithm given by $\log(z)=\log(|z|)+i\arg(z)$.

Here, we have

$$\begin{align} (-1)^\sqrt2&=e^{\sqrt2\log(-1)}\\\\ &=e^{\sqrt2 (i(2n+1)\pi)}\\\\ &=e^{i(2n+1)\pi \sqrt2} \\\\ &=\cos((2n+1)\pi \sqrt2)+i\sin((2n+1)\pi\sqrt2) \end{align}$$

where $n\in \mathbb{N}$.

If we choose the principal branch of the logarithm (i.e., $n=0$), then $-1=e^{i\pi}$ and we have

$$(-1)^\sqrt2=\cos(\pi\sqrt2)+i\sin(\pi\sqrt2)$$

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Now, if $p$ is an odd integer and $q$ is an even integer, then we have

$$\begin{align} (-1)^{p/q}&=e^{(p/q)\log(-1)}\\\\ &=e^{(p/q) (i(2n+1)\pi)}\\\\ &=e^{i(2n+1)p\pi /q} \\\\ &=\cos((2n+1)p\pi/q )+i\sin((2n+1)p \pi/q ) \end{align}$$

If $p$ is odd, then certainly $p(2n+1)$ is also odd and $p(2n+1)/q$ is not an integer. Therefore, $\sin((2n+1)p\pi/q)\ne 0$ and $(-1)^{p/q}$ has a non-zero imaginary part.

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By continuity of the complex exponential function, for any $\epsilon>0$, there exists a $\delta>0$ such that

$$\left|e^{i(2n+1)p\pi/q}-e^{i(2n+1)\sqrt 2\,\pi}\right|<\epsilon$$

whenever $|(2n+1)p\pi /q-(2n+1)\sqrt2 \pi|<\delta$.

By the density of the rational numbers, for this $\delta>0$ there exist integers $(2n+1)p$ and $q$ such that $|(2n+1)p\pi /q-(2n+1)\sqrt2 \pi|<\delta$.

And we are done!

Answer 2

As Mark pointed out in the comment,

$$(-1)^{\sqrt{2}}=(i^2)^{\sqrt{2}}=i^{2\sqrt{2}}\text{.}$$

Now we know that

$$i^{2\sqrt{2}}=(e^{\pi i/2})^{2\sqrt{2}}=e^{\pi i\sqrt{2}}\text{,}$$

and by De Moivre's formula,

$$e^{\pi i\sqrt{2}}=\cos(\pi\sqrt{2})+i\sin(\pi\sqrt{2})$$

which is complex (and transcendental).

Not sure how to use your given fact, though, since $\sqrt{2}$ is irrational.