# 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\$.