1
$\begingroup$

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

I have tried adding the first two and the second two separately, then adding those sums but how do I get a numerical value as the answer?

Thanks

$\endgroup$
4
$\begingroup$

HINT : $$\frac{\omega}{1-\omega^2}+\frac{\omega^2}{1-\omega^4}+\frac{\omega^3}{1-\omega}+\frac{\omega^4}{1-\omega^3}$$ $$=\frac{\omega}{1-\omega^2}+\frac{\omega^2}{1-\omega^4}+\frac{\omega^7}{\omega^4-\omega^5}+\frac{\omega^6}{\omega^2-\omega^5}$$

$\endgroup$
1
$\begingroup$

The sum of the first and fourth terms is \begin{align*} \frac{\omega}{1 - \omega^2} + \frac{\omega^4}{1 - \omega^3} &= \frac{\omega (1 - \omega^3) + \omega^4 (1 - \omega^2)}{(1 - \omega^2)(1 - \omega^3)} \\ &= \frac{\omega - \omega^4 + \omega^4 - \omega^6}{(1 - \omega^2)(1 - \omega^3)} \\ &= \frac{\omega - \omega^4 + \omega^4 - \omega}{(1 - \omega^2)(1 - \omega^3)} \\ &= 0, \end{align*}and the sum of the second and third terms is \begin{align*} \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} &= \frac{\omega^2 (1 - \omega) + \omega^3 (1 - \omega^4)}{(1 - \omega^4)(1 - \omega)} \\ &= \frac{\omega^2 - \omega^3 + \omega^3 - \omega^7}{(1 - \omega^4)(1 - \omega)} \\ &= \frac{\omega^2 - \omega^3 + \omega^3 - \omega^2}{(1 - \omega^4)(1 - \omega)} \\ &= 0. \end{align*}Therefore, the sum of all four terms is $\boxed{0}$.

$\endgroup$
0
$\begingroup$

Maybe more direct $$ \frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3} = \frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^{-1}} + \frac{\omega^{3}}{1 - \omega} + \frac{\omega^{-1}}{1 - \omega^{-2}} $$

$\endgroup$
0
$\begingroup$

A less efficient answer with a more general approach.


You may notice that $\{\omega,\omega^2,\omega^3,\omega^4\}$ are the roots of $\frac{x^5-1}{x-1}$.
If we set $Z=\{\omega,\omega^2,\omega^3,\omega^4\}$ we have

$$ \sum_{z\in Z}\frac{z}{1-z^2}=\sum_{z\in Z}\frac{z^3}{1-z}=\sum_{z\in Z}\frac{1}{1-z}-\sum_{z\in Z}(1+z+z^2)=-2+\sum_{z\in Z}\frac{1}{1-z}.$$ If $z\in Z$, $1-z$ is a root of $\frac{1-(1-x)^5}{x}=x^4-5x^3+10x^2-10x+5$.
By Vieta's theorem it follows that $$ \sum_{z\in Z}\frac{1}{1-z} = \frac{10}{5} = 2$$ hence: $$ \sum_{z\in Z}\frac{z}{1-z^2} = \color{red}{0}.$$ Key steps:

  1. $z\mapsto z^3$ is a bijection on $Z$
  2. for any $k\in[1,4]$ we have $\sum_{z\in Z}z^k = -1$.
$\endgroup$

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.