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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 combining the first and third terms & first and last terms. Here is what I have so far:

\begin{align*} \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^4}{1 + \omega^3} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} \\ &= \dfrac{\omega(1+\omega^3) + \omega^4(1+\omega^2)}{(1+\omega^2)(1+\omega^3)} + \dfrac{\omega^2(1+\omega) + \omega^3(1+\omega^4)}{(1+\omega^4)(1+\omega)} \\ &= \dfrac{\omega + 2\omega^4 +\omega^6}{1+\omega^2 + \omega^3 + \omega^5} + \dfrac{\omega^2 + 2\omega^3 + \omega^7}{1+\omega + \omega^4 + \omega^5} \\ &= \dfrac{2\omega + 2\omega^4}{2+\omega^2 + \omega^3} + \dfrac{2\omega^2 + 2\omega^3}{2+\omega+\omega^4} \end{align*}

OR

\begin{align*} \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^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3} + \frac{\omega^2}{1 + \omega^4} \\ &= \dfrac{\omega(1+\omega) + \omega^3(1+\omega^2)}{(1+\omega)(1+\omega^2)} + \dfrac{\omega^2(1+\omega^3) + \omega^4(1+\omega^4)}{(1+\omega^3)(1+\omega^4)} \\ &= \dfrac{\omega + \omega^2 + \omega^3 + \omega^5}{1+\omega + \omega^2 + \omega^3} + \dfrac{\omega^2 + \omega^4 + \omega^5 + \omega^8}{1 + \omega^3 + \omega^4 + \omega^7} \\ &= \dfrac{2\omega+\omega^2+\omega^3}{1+\omega+\omega^2+\omega^4} + \dfrac{1+\omega+\omega^2+\omega^4}{1+2\omega^3+\omega^4} \end{align*}

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    $\begingroup$ Please note that this is different from previous similar questions because of the denominators of the terms in the original question. Their signs are flipped. Thanks! $\endgroup$ – Ojasw Upadhyay Aug 18 '18 at 6:08
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Expand 2nd and 4th fraction with $\omega $ and $\omega ^2$ respectively: $$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}=\frac{\omega}{1 + \omega^2} + \frac{\omega^3}{\omega+ 1} + \frac{\omega^3}{1 + \omega} + \frac{\omega}{\omega^2+1}$$

$$=2\frac{\omega}{1 + \omega^2} + 2\frac{\omega^3}{\omega+ 1} $$ $$=2\frac{\omega^2+\omega + \omega^3+1}{(\omega^2+1)(\omega+ 1)}=2 $$

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  • $\begingroup$ @henningmakholm nice $\endgroup$ – user2661923 Aug 18 '18 at 22:31
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Alt. hint:   let $\,z=\omega+\dfrac{1}{\omega}\,$ so that $\,z^2=\omega^2+\dfrac{1}{\omega^2}+2\,$, then use that $\,\omega^4=\bar\omega\,$ and $\,\omega^3=\bar\omega^2\,$ so the sum is:

$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\bar\omega^2}{1 + \bar\omega^4} + \frac{\bar\omega}{1 + \bar\omega^2} = 2 \operatorname{Re}\left(\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} \right) = 2 \operatorname{Re}\left(\frac{1}{z} + \frac{1}{z^2-2}\right)$$

But $\,0=\omega^5-1=(\omega-1)\left(\omega^4+\omega^3+\omega^2+\omega+1\right)=\omega^2(\omega-1)\left(z^2 + z - 1\right)\,$, so $\,z^2+z-1=0\,$ and:

$$\require{cancel} \frac{1}{z} + \frac{1}{z^2-2} = \frac{1}{z}+\frac{1}{-z-1} = \frac{\cancel{-z}-1+\cancel{z}}{-z^2-z} = \frac{-1}{-1} $$

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