Solving the equation $z^7=-1$ 
Solve the equation $z^7=-1$

My attempt:
$$z=x+yi$$
$$(x+yi)^7+1=0$$
$$(x^2+2yi-y^2)^3(x+yi)+1=0$$
but now it's start to look ugly.
I'm sure that there is a simple way
 A: We can say that $$z^7=-1$$
or,$$z^7=\cos\pi+i\sin\pi$$
or,$$z^7=\cos(4n+2)\pi+i\sin(4n+2)\pi \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$ 
or,$$z=\left[\cos(4n+2)\pi+i\sin(4n+2)\pi\right]^{\frac{1}{7}} \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$
or,$$z=\cos(\frac{4n+2}{7})\pi+i\sin(\frac{4n+2}{7})\pi \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$
A: $$z^7=-1\Longleftrightarrow$$
$$z^7=|-1|e^{\arg(-1)i}\Longleftrightarrow$$
$$z^7=1\cdot e^{\pi i}\Longleftrightarrow$$
$$z^7=e^{\pi i}\Longleftrightarrow$$
$$z=\left(e^{\left(\pi+2\pi k\right)i}\right)^{\frac{1}{7}}\Longleftrightarrow$$
$$z=e^{\frac{1}{7}\left(\pi+2\pi k\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-6$

So the solutions are:
$$z_0=e^{\frac{1}{7}\left(\pi+2\pi\cdot 0\right)i}=e^{\frac{\pi}{7}i}$$
$$z_1=e^{\frac{1}{7}\left(\pi+2\pi\cdot 1\right)i}=e^{\frac{3\pi}{7}i}$$
$$z_2=e^{\frac{1}{7}\left(\pi+2\pi\cdot 2\right)i}=e^{\frac{5\pi}{7}i}$$
$$z_3=e^{\frac{1}{7}\left(\pi+2\pi\cdot 3\right)i}=e^{\pi i}=-1$$
$$z_4=e^{\frac{1}{7}\left(\pi+2\pi\cdot 4\right)i}=e^{-\frac{5\pi}{7}i}$$
$$z_5=e^{\frac{1}{7}\left(\pi+2\pi\cdot 5\right)i}=e^{-\frac{3\pi}{7}i}$$
$$z_6=e^{\frac{1}{7}\left(\pi+2\pi\cdot 6\right)i}=e^{-\frac{\pi}{7}i}$$
A: HINT:
Try using this,(one of the solution)

$$
z^7=e^{i\pi}
\\z=e^{i\frac{\pi}{7}}
$$
