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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 8, 2020

Best Answer 

 #2
avatar+24054 
+2

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

 

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

 

laugh

 Jan 8, 2020
 #1
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The answer is -2.

 Jan 8, 2020
 #2
avatar+24054 
+2
Best Answer

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

 

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

 

laugh

heureka Jan 8, 2020
 #3
avatar+107099 
+1

Very nice, heureka    !!!!

 

 

cool cool cool

CPhill  Jan 8, 2020
 #4
avatar+24054 
+2

Thank you, CPhill !

 

laugh

heureka  Jan 8, 2020

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