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.
Thank you in advance
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2 Answers
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5
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 $q$.
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$\begingroup$ 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. $\endgroup$– CoinTossCommented Oct 22, 2014 at 14:37
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1$\begingroup$ @CoinToss: No. You should write it in polar form. $\endgroup$– LucianCommented Oct 22, 2014 at 14:38
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$\begingroup$ Then you should use the polar notation $q^{5}=re^{i\theta}$ $\endgroup$– MartiganCommented Oct 22, 2014 at 14:39
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$\begingroup$ 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? $\endgroup$– CoinTossCommented Oct 22, 2014 at 14:43
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1$\begingroup$ @CoinToss $z=1+i=\sqrt{2}e^{i\frac{\pi}{4}}$... $\endgroup$– MartiganCommented Oct 22, 2014 at 14:46
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Hint:
Substitute $q^5$ with $r$, so you get the equation $r^2-2r+2=0$
