Find all complex solutions $z^{10} + 2z^5 + 2 = 0$. So I’m working on this equation $z^{10} + 2z^5 + 2 = 0$ to find all complex solutions, and I think I managed to solve it, but I can’t find solution manual for it, since it is really old exam task. The thing that makes me uncomfortable with my solution is that, shouldn’t I get just 10 solutions? But when I put in all $k$ values($k = 0,1,2,3,4$), you get 12 different angle solutions. Isn’t that wrong?
My answer:
$$ \sqrt{\mathstrut 2}^{1/5}e^{\frac{\left(\pm\frac\pi4i+2\pi k \right)}5} $$
Sorry could't figure out how to put 5 in denominator of the polar formula...
 A: solve for $$w^2 + 2w + 2=0$$
which gives $$w_{1,2} = -1 \pm i = \sqrt{2} e^{i(\pi \pm \frac{\pi}{4})}$$
Now you got two equations to solve 
\begin{align}
z_1^5 &= \sqrt{2} e^{i(\pi + \frac{\pi}{4})} \\
z_2^5 &= \sqrt{2} e^{i(\pi - \frac{\pi}{4})} 
\end{align}
which gives
\begin{align}
z_1 &= \sqrt{2} e^{i(\pi + \frac{\pi}{4} + \frac{2k\pi}{5})} \\
z_2 &= \sqrt{2} e^{i(\pi - \frac{\pi}{4}+ \frac{2k\pi}{5})} 
\end{align}
for $k \in \lbrace 0,1,2,3,4 \rbrace$
A: $$1+z^5=-1\pm i=\sqrt2e^{i(2n\pi+\pi\pm\pi/4)}$$  where $n$ is any integer
$$z=2^{1/10}\text{exp}\left(\dfrac{i\left(2n\pi+\pi\pm\pi/4\right)}5\right)$$
where $0\le n\le4$
Observe that  $$\text{exp}\left(\dfrac{i\left(2n\pi+\pi+\pi/4\right)}5\right)\ne\text{exp}\left(\dfrac{i\left(2n\pi+\pi-\pi/4\right)}5\right)$$  as the equality needs $\text{exp}\left(\dfrac{i\pi/2}5\right)=1$
Like How to solve $x^3=-1$?, we can prove that  there is no repeated root of the given equation 
Now if for $n_1<n_2$ $$\text{exp}\left(\dfrac{i\left(2n_1\pi+\pi+\pi/4\right)}5\right)=\text{exp}\left(\dfrac{i\left(2n_2\pi+\pi+\pi/4\right)}5\right)$$
$\iff\dfrac{2(n_1-n_2)}5$ is a multiple of $2\pi$
$\iff n_1-n_2$ is a multiple of $5$ which is not possible as $0\le n_1<n_2\le4$
Similarly for 
$$\text{exp}\left(\dfrac{i\left(2n_1\pi+\pi-\pi/4\right)}5\right)=\text{exp}\left(\dfrac{i\left(2n_2\pi+\pi-\pi/4\right)}5\right)$$
