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)
 A: Welcome to Math stack! We have $$u^2+2u+2$$
Using your assumptions. We now get $u=-1+i$ & $ u=-1-i$
For the solution of u.
We have $u=z^{5}$
$$-1+5i=z^{5}$$
And
$$-1-5i=z^{5}$$
let &$z=x+yi$
Then, the argument becomes $tan^{-}(\frac{y}{x})$ thus $z^{5}$ has an argument $5*tan^{-}(\frac{y}{x})$
The argument of $-1+5i$ is $tan^{-}(5/-1)$, which is $-78.69^{o}+2k\pi$
We have an equation,
$$-78.69^{\circ}+2k*180^{\circ}=5*tan^{-}(\frac{y}{x})$$
We also have $$x^{2}+y^{2}=26^{\frac{1}{5}}$$
Try also with the other complex number and try to solve!
We have two equations and I think they can be solvable, 

You can also use $z=(1-5i)^{\frac{1}{5}}$ and change it to polar form.
The magnitude of z:
$$r=26^{1/5}$$
$$\theta=\frac{tan^{-}(-5)}{5}+2k\pi$$
Where k is any natural number.The same for the second complex number.
A: 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 $-1-i=\sqrt{2}e^{-i\frac{3\pi}{4}}$.
$z_1=\sqrt[10]{2}e^{i\frac{3\pi}{20}}=\sqrt[10]{2}\left[\cos\left(\frac{3\pi}{20}\right)+i\sin\left(\frac{3\pi}{20}\right)\right]\approx 0.954857+0.486575i$
$z_2=\sqrt[10]{2}e^{i\frac{7\pi}{20}}=\sqrt[10]{2}\left[\cos\left(\frac{7\pi}{20}\right)+i\sin\left(\frac{7\pi}{20}\right)\right]\approx 0.486575+0.954957i$
$\vdots$
