0
$\begingroup$

(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!

$\endgroup$
1
$\begingroup$

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

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

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

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:

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.