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

 Jan 10, 2020
 #1
avatar+108771 
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\(\theta=\frac{2\pi}{5}\)

 

\(w=e^{(2k\pi/5)i}\qquad k=1,2,3,4\)

 

That is a start.

 Jan 10, 2020
 #2
avatar+24422 
+1

Let \(\omega\) be a complex number such that \(\omega^5 = 1\) and \(\omega \neq 1\).

Compute \(\dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^2}{1 - \omega^4} + \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^4}{1 - \omega^3}\)

 

 

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^2}{1 - \omega^4} + \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^4}{1 - \omega^3} } \quad &| \quad \omega^4 = \dfrac{\omega^5}{\omega}=\dfrac{1}{\omega} \\\\ &=& \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^2}{1 - \dfrac{1}{\omega}} + \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^4}{1 - \omega^3} \\\\ &=& \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^2}{\dfrac{\omega-1}{\omega}} + \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^4}{1 - \omega^3} \\\\ &=& \dfrac{\omega}{1 - \omega^2} - \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^4}{1 - \omega^3} \\\\ &=& \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^4}{1 - \omega^3} \quad &| \quad \omega^3 = \dfrac{\omega^5}{\omega^2}=\dfrac{1}{\omega^2} \\\\ &=& \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^4}{1 - \dfrac{1}{\omega^2}} \\\\ &=& \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^4}{\dfrac{\omega^2-1}{\omega^2}} \\\\ &=& \dfrac{\omega}{1 - \omega^2} - \dfrac{\omega^6}{1 - \omega^2} \quad | \quad \omega^6 = \omega^5 \omega=\omega \\\\ &=& \dfrac{\omega}{1 - \omega^2} - \dfrac{\omega}{1 - \omega^2} \\\\ &=& \mathbf{0} \\ \hline \end{array}\)

 

laugh

 Jan 10, 2020
edited by heureka  Jan 10, 2020
edited by heureka  Jan 10, 2020

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