# How to solve higher grade polynomials of complex numbers $q^{10}-2q^5+2=0$

If I wanted to find the roots for $q^{10}-2q^5+2=0$, how would I go about doing that?

I tried treating it like a quadratic equation, but couldn't get there. I also tried putting $q=(a+ib)$ but that didn't do much.

You have to solve it in two steps.

Let $x=q^{5}$.

Then $x²-2x+2=0$

Solve this, and then solve the first equation to find the solutions in $q$.

• I should've added that I tried this, I got the following from it $z^5 = 1 \pm i$ but I don't know how to go from there. Should I expand the $q^5$ as $(a+ib)^5$ through the binomial theorem to then analyse all factors? Seems like very much work. Commented Oct 22, 2014 at 14:37
• @CoinToss: No. You should write it in polar form. Commented Oct 22, 2014 at 14:38
• Then you should use the polar notation $q^{5}=re^{i\theta}$ Commented Oct 22, 2014 at 14:39
• So I'd get $$q = \frac{1}{5}\cdot e^{i\frac{1}{5}\cdot\frac{\pi}{4}} = \frac{1}{5}(\cos{(2\pi\cdot k+\frac{\pi}{20})}+i\cdot\sin{(2\pi\cdot k+\frac{\pi}{20})}),\quad 0 \le k \le 4$$ And then for each solution, the conjugate would be another solution? Commented Oct 22, 2014 at 14:43
• @CoinToss $z=1+i=\sqrt{2}e^{i\frac{\pi}{4}}$... Commented Oct 22, 2014 at 14:46

Hint:

Substitute $q^5$ with $r$, so you get the equation $r^2-2r+2=0$