I started to think about the function $f(x) =x^{1/x} $ and how it behaves. How would you show for which negative real values of x that $f(x)$ gives real values?
I tried by rewritting, set $y=-x$ and we get $\displaystyle-\frac{(-1)^{((y-1)/y)} }{y^{(1/y)} } $. The nominator will be complex if $y>1$. Is there any other way to show this? By using euler's formula maybe? Since it is basically just the finding the roots for all negative x.
On
The expression $x^{1/x}$ would always yield a real value for any $x<0$. To see so, let $x=-a$, with $a>0$,
$$x^{1/x}=(-a)^{-1/a} =(-a)^{2(-\frac1{2a})}=(a^2)^{-\frac1{2a}} =\frac1{(a^2)^{\frac1{2a}}}=\frac1{a^{\frac1{a}}}$$
However, the real value obtained above may not be unique. For example, if $a$ is an odd natural number, that is, $a=2k+1$, the expression $x^{1/x}$ could also lead to,
$$x^{1/x}=[-(2k+1)]^{-1/(2k+1)} = \frac1{[-(2k+1)]^{1/(2k+1)} } =-\frac1{(2k+1)^{1/(2k+1)} } = -\frac1{a^{\frac1{a}}} $$
The reason is that $f(x) = x^{1/x}$ is undefined for $x<0$.
This is real when $x$ is negative only when $x$ is also an odd integer or a rational with odd numerator, for if $x$ were an even integer, we have complex values. If $x$ is rational, then it should have the form $$\frac{\text{odd integer}}{\text{nonzero integer}},$$ for the same reason as in the integer case. If $x$ is irrational, then we might want to write the expression as $$e^{\log x^{1/x}}=e^{\frac{\log x}{x}},$$ which is not then real.
Thus, in summary, if $x<0,$ then it should either be an odd integer, or a rational with odd numerator as well.