Solve Complex Equation with several terms I have a complex number $z = 3 + 3i$
And I want to find all solutions of
$z^{10} + 2z^{5} + 2 = 0$
I'm kinda lost. I recognise the fact that I can substitute 
$u = z^{5}$ and rewrite the equation as $u^{2} + 2u + 2 = 0$ and take it from there, but I'm lost and dont have a solution guide and thus wonder if I can get a hint here! Would be much appreciated. Thanks in advance (My first post here btw)

Hint:
Since $-1+i$ and $-1-i$ are solutions of $u^2+2u+2=0$; then, in order to solve $z^{10}+2z^5+2=0$ we need find the $5$th roots of $-1+i=\sqrt{2}e^{i\frac{3\pi}{4}}$ and $-1-i=\sqrt{2}e^{-i\frac{3\pi}{4}}$.
$z_1=\sqrt[10]{2}e^{i\frac{3\pi}{20}}=\sqrt[10]{2}\left[\cos\left(\frac{3\pi}{20}\right)+i\sin\left(\frac{3\pi}{20}\right)\right]\approx 0.954857+0.486575i$
$z_2=\sqrt[10]{2}e^{i\frac{7\pi}{20}}=\sqrt[10]{2}\left[\cos\left(\frac{7\pi}{20}\right)+i\sin\left(\frac{7\pi}{20}\right)\right]\approx 0.486575+0.954957i$
$\vdots$
