How do I calculate this kind of expressions $(-1)^{-\frac{2}{5}}$? I used a tool and found that it is equal to $0.31 - 0.95i$
Can someone please tell me the way I reach this result step by step?
 A: 
Taking the principal value of the logarithm we obtain
  \begin{align*}
\color{blue}{(-1)^{-\frac{2}{5}}}&=e^{-\frac{2}{5}\mathrm{Log}(-1)}\\
&=e^{-\frac{2}{5}\left(\ln|-1|+i\mathrm{Arg(-1)}\right)}\\
&=e^{-\frac{2}{5}\left(0+i\pi\right)}=e^{-\frac{2i\pi}{5}}\\
&\,\,\color{blue}{=\cos\left(\frac{2\pi}{5}\right)-i\sin\left(\frac{2\pi}{5}\right)}\\
&\,\,\color{blue}{\doteq0,31-0,95i}
\end{align*}

A: Note that  $$(-1)^{-2/5}=\frac1{\sqrt[5]{(-1)^2}}=\frac1{\sqrt[5]1}$$ so on the complex plane there will be $5$ solutions.
Write $$1=\cos0+i\sin0$$ and use De Moivre's generalisation formula.
A: That's only one possible value.
$(-1)^{-\frac 25}$ will be a number $z$ so that $z^5 = (-1)^{-2} = 1$.  $z = 1$ is one such number.  $z = 1$ is the only real such number.  There are four other such complex numbers.
If we view a non-zero complex number, $a + bi$ as a point $(a,b)$ in a plane the point has a distinct distance $r = \sqrt{a^2 + b^2} > 0$ from $(0,0)$.  And the angle between the real axis, the origin, and the point is a distcint angle $\theta$ were $\cos \theta = \frac {a}r$ and $\sin \theta = \frac {b}r$.
So the complex number can be written as $r(\cos \theta + i\sin \theta)$.
And here's the astonishing  thing about it:  If $z = r(\cos \theta + i\sin \theta)$ and $w = s(\cos \eta + i\sin \eta)$ then $z\cdot w = rs(\cos(\theta + \eta) + i \sin (\theta + \eta)$.
[Just do the trig:  $z\cdot w = rs(\cos \theta + i\sin \theta)(\cos \eta + i\sin \eta)$ and $(\cos \theta + i\sin \theta)(\cos \eta + i\sin \eta) = (\cos \theta\cos \theta - \sin \theta\sin \eta)+ i(\sin\theta \cos \eta + \cos \theta \sin \eta)$ and $\cos (\theta + \eta) + i\sin(\theta + \eta) = (\cos \theta\cos \theta - \sin \theta\sin \eta)+ i(\sin\theta \cos \eta + \cos \theta \sin \eta)$]
So that means $z^k = r^k(\cos k\theta + i\sin k \theta)= r^k(\cos k(\theta+ 2m\pi) + i\sin k (\theta+ 2m\pi))$.
And that means $z^{\frac 1k} = r^{\frac 1k}(\cos (\frac {\theta}k + m*\frac {2\pi}k) + i\sin (\frac {\theta}k + m*\frac {2\pi}k))$.
So if $z^5 = 1= \cos 0  + i \sin 0$ then $z = \cos {m \frac {2\pi}5} + i \sin{m\frac {2\pi}5}$.
$0.31 - 0.95i\approx \cos \frac {8\pi}5 + i\sin \frac {8\pi}5$.  i.e. the solution when $m = 4$.  The other four are $\cos \frac {2\pi}5 + i\sin \frac {2\pi}5\approx 0.31 + 0.95i$ and $\cos \frac {4\pi}5 + i\sin \frac {4\pi}5 \approx -0.81 + 0.59i$ and $\cos \frac {6\pi}5 + i\sin \frac {6\pi}5 \approx -0.81 - 0.59i$ and $\cos 0 + i\sin 0 =1$
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When we get comfortable with these ideas we define
$e^{\theta i} :=\cos \theta + i\sin \theta $ (because that satisfies the property that $\frac {d e^z}{d z} = e^z$).
From there we can classify all complex numbers as $r*e^{i\theta}$.  This allows us to derive the famous Euler's identity $e^{\pi i} = \cos \pi + i \sin \pi = -1$.
As $z^5 = 1$ has five solutions which do we declare to be the fifth root.  Well.... the real root is $1$ of course but as the expressing was $(-1)^{-\frac 25}$ and fractiona powers of negative numbers implies complex roots and for one reason or another the concept of "principal root" and "principal" Logarithms leads to Markus Shauer's answer.
A: Use $-1 = e^{i\pi +i2\pi n}$ for $n\in\mathbb{Z}$. Note that we need the factor of unity $e^{i2\pi n}$ in order to discover all solutions:
\begin{align}
(-1)^{-\frac{2}{5}} &= \left(e^{i\pi(1+2 n)}\right)^{-\frac{2}{5}} 
= e^{-i\frac{2\pi}{5}(1+2n)} 
\\
&= \cos\left(-\frac{2\pi}{5}(1+2n)\right) + i \sin\left(-\frac{2\pi}{5}(1+2n)\right)
\\
&= \cos\left(\frac{2\pi}{5}(1+2n)\right) - i \sin\left(\frac{2\pi}{5}(1+2n)\right)
\end{align}
There are only five distinct solutions with $n=0,1,2,3,4$. At $n=5$, the solutions repeat. This is because you have a fifth root. It appears that the $n =0\mod 5$ solution is the one you referenced in your original post.
