complex conjugate pairs of a quartic

Question

I tried my hand at this question, which included finding the partial fractions of $\frac{x^2}{1-x^5}$. I found a factor of $1-x$ for the denominator, but I do not know how to work out the complex conjugate pairs of $x^4+x^3+x^2+x+1$ as M. Strochyk gave them. I could probably do it if one root was given.

Answer 1

What M. Strochyk did was solve $1-x^5 = 0$ for real and complex roots using De Moivre's Theorem. First, we want to solve: $$x^5 = 1$$ We first convert the $"1"$ into complex form, which is $1+0i$. Then we convert $1+0i$ into polar form, and we need to find $r$ and $\theta$. $r = \sqrt{1^2+0^2} = 1$ and $\theta = \arctan(\frac{0}{1}) = 0, \pi, 2\pi,... $

So which is $1(\cos(0 + n\pi) + i\sin(0 + n\pi))$ for all integers $n > 0$. So: $$x^5 = 1(\cos(0 + n\pi) + i\sin(0 + n\pi))$$ $$x = (\cos(0 + n\pi) + i\sin(0 + n\pi))^{1/5}$$ Using De Moivre's Theorem: $$x = (\cos((0 + n\pi)*(1/5)) + i\sin((0 + n\pi)*(1/5))$$ We know there are a maximum of $5$ roots, therefore we want $n$ to be $0,1,2,3$ and $4$. Then you plug $n$ into the equation above, and you'll get M.Strochyk's answers.