# Roots of Unity, Precalculus

(a) Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find $$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}$$

(b)Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find $$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}$$

For these two problems, I have tried to factor the denominators and rearrange the fractions and adding them up.

Any help is appreciated!

Hint: (a) We may write $$\frac{\omega}{1 + \omega^2}\frac{\omega^4}{\omega^4} = \frac{1}{\omega^4 + \omega}$$

then \begin{align}\frac{1}{\omega^4 + \omega} + \frac{1}{\omega^3 + \omega^2} + \frac{1}{\omega^3 + \omega^2} + \frac{1}{\omega^4 + \omega} &= \frac{2}{\omega^4 + \omega} + \frac{2}{\omega^3 + \omega^2} \\&= 2 \Bigg[\frac{1 + \omega + \omega^2 + \omega^3 + \omega^4 -1 }{(\omega^4 + \omega)(\omega^3 + \omega ^2) }\Bigg] \\&=-\frac{2}{(1 + \omega)^2} \end{align}

• What would you suggest be the next step? – Math is Life Apr 27 '15 at 0:09
• The answer was given according to what the problem has presented. So there isn't a next step. Next step is letter (b). – Aaron Maroja Apr 27 '15 at 0:36
• Ah, I see. I assumed that there would be a numerical value for letter (a), which is why I asked. – Math is Life Apr 27 '15 at 0:37
• There could be, if you chose a $\omega$, no problem. – Aaron Maroja Apr 27 '15 at 1:21
• Where $\omega$ can be any complex number? – Math is Life Apr 27 '15 at 1:27

Note that $\omega^k$ and $\omega^{5-k}$ are conjugate, and that $\omega^5=1$. You can use this facts to simplify the expressions.

Something like this:

\begin{align} \frac\omega{1+\omega^2}+\frac{\omega^4}{1+\omega^3}&=\frac{2(\omega+\omega^4)}{2+\omega^2+\omega^3}=\frac{4\cos72^o}{2+2\cos144^o}\\ \\ &=\frac{4\cos72^o}{2+2\cos^272^o-2\sin^272^o}=\frac1{\cos72^o} \end{align}

i just wanna show first sum has a good property:

Let $w\neq 1$, then $w^4+w^3+w^2+w+1=0$.

Lets make some steps:

$w^2((w+\frac1w)^2+(w+\frac1w)-1)=0$. Say $w+\frac 1w=\frac{1+w^2}{w}=\frac1u$ (here $u=\frac {w}{1+w^2}$). Then

$u^2-u-1=0$. Here sum of roots is $1$.

Now calculate $\sum\limits_{x^5=1,x \neq 1}\frac{x}{1+x^2}$