When is $displaystyle {(x)}^{alpha }=-{(x)}^{alpha}$, for $x<0$ and $0

Let $x$ be a negative real number and $0<alpha <1$, I would like to know why ${(x)}^{alpha }$ is not a negative real number.

For example wolfram alpha shows that
$$
{(-5)}^{0.33 }=0.856097+1.47919i
$$
but it’s not $-1.70906$.

My question:

When is it true that
$$
{(x)}^{alpha }=-{(x)}^{alpha},
$$ for $x<0$ and $0<alpha <1$ ?

Answer

$(x)^alpha=-(x)^alpha$ does not make sense since this implies that $x^alpha=0$. You are probably asking if $(x)^alpha=-(-x)^alpha$ is true in general for $x<0$ and $0<alpha<1$.

The answer is NO.

For $x<0$ and $0<alpha<1$, one has to refers to complex analysis in order to make sense of $x^alpha$. If $alpha=frac{n}{m}$ for some integers $m$ and $n$, then
essentially you need to know the $m$-th root of unity. If otherwise $alpha$ is irrational, then
$$
x^alpha:=e^{alphalog x}
$$
and the right hand side is a long story.


[Added:] $x^alpha=-(-x)^alpha$ for $x<0$ and $0<alpha<1$ could be true if one stricts oneself to the real setting and further assumes that $alpha=dfrac{1}{m}$ where $m$ is an odd interger. For instance, you do have
$$
(-2)^{1/3}=-(2)^{1/3}
$$
where $y=(-2)^{1/3}$ is defined as a real number such that $y^3=-2$.

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