# 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)

• Are you trying to solve $z^{10}+2z^5+2=0$? (If so, your substitution is useful, and you can apply the quadratic formula.) What does $z=3+3i$ have to do with this? Commented Aug 23, 2015 at 19:06
• Are you trying to find another complex number to fit this solution or is z defined and you put z in the equation and get another complex number? Commented Aug 23, 2015 at 19:23
• thanks. I was confused with the $z$ from part A of the problem Commented Aug 24, 2015 at 18:13

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.

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 $-1-i=\sqrt{2}e^{-i\frac{3\pi}{4}}$.

$z_1=\sqrt[10]{2}e^{i\frac{3\pi}{20}}=\sqrt[10]{2}\left[\cos\left(\frac{3\pi}{20}\right)+i\sin\left(\frac{3\pi}{20}\right)\right]\approx 0.954857+0.486575i$

$z_2=\sqrt[10]{2}e^{i\frac{7\pi}{20}}=\sqrt[10]{2}\left[\cos\left(\frac{7\pi}{20}\right)+i\sin\left(\frac{7\pi}{20}\right)\right]\approx 0.486575+0.954957i$

$\vdots$