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Simplify \(\left( \frac{3 + i \sqrt{3}}{2} \right)^8 + \left( \frac{3 - i \sqrt{3}}{2} \right)^8\)(\(i\) is imaginary)

 Aug 18, 2019
 #1
avatar+23884 
+3

Simplify
\(\left( \dfrac{3 + i \sqrt{3}}{2} \right)^8 + \left( \dfrac{3 - i \sqrt{3}}{2} \right)^8\)( \(i\) is imaginary)

 

\(\text{Let $z = \dfrac{3 + i \sqrt{3}}{2}$} \\ \text{Let $\overline{z} = \dfrac{3 - i \sqrt{3}}{2}$} \)

 

\(\begin{array}{|rcll|} \hline z\cdot \overline{z} &=& \left(\dfrac{3}{2}\right)^2 +\left( \dfrac{\sqrt{3}}{2} \right)^2 \\ &=& \dfrac{9}{4} + \dfrac{3}{4} \\ \mathbf{z\cdot \overline{z}} &=& \mathbf{3} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \left(z+\overline{z} \right) &=& \dfrac{3}{2} + \dfrac{i \sqrt{3}}{2}+ \dfrac{3}{2} - \dfrac{i \sqrt{3}}{2} \\ &=& 2\cdot \dfrac{3}{2} \\ \mathbf{\left(z+\overline{z} \right)} &=& \mathbf{3} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \left(z+\overline{z} \right)^2 = 3^2 &=& z^2+2\cdot z\cdot \overline{z} + \overline{z}^2 \\ 3^2 &=& z^2 + \overline{z}^2 +2\cdot 3 \\ z^2 + \overline{z}^2 &=& 9-6 \\ \mathbf{z^2 + \overline{z}^2} &=& \mathbf{3} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \left(z^2+\overline{z}^2 \right)^2 = 3^2 &=& z^4+2\cdot z^2\cdot \overline{z}^2 + \overline{z}^4 \\ 9 &=& z^4+ \overline{z}^4 + 2\cdot \left(z\cdot \overline{z}\right)^2 \\ 9 &=& z^4+ \overline{z}^4 + 2\cdot 3^2 \\ z^4 + \overline{z}^4 &=& 9-18 \\ \mathbf{z^4 + \overline{z}^4} &=& \mathbf{-9} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \left(z^4+\overline{z}^4 \right)^2 = (-9)^2 &=& z^8+2\cdot z^4\cdot \overline{z}^4 + \overline{z}^8 \\ 81 &=& z^8+ \overline{z}^8 + 2\cdot \left(z\cdot \overline{z}\right)^4 \\ 81 &=& z^8+ \overline{z}^8 + 2\cdot 3^4 \\ z^8 + \overline{z}^8 &=& 81-2\cdot 81 \\ \mathbf{z^8 + \overline{z}^8} &=& \mathbf{-81} \\ \hline \end{array} \)

 

\(\left( \dfrac{3 + i \sqrt{3}}{2} \right)^8 + \left( \dfrac{3 - i \sqrt{3}}{2} \right)^8 = \mathbf{z^8 + \overline{z}^8} = \mathbf{-81} \)

 

laugh

 Aug 18, 2019
edited by heureka  Aug 19, 2019
 #2
avatar+6046 
+3

another method

 

\(z = r e^{i\theta}\\ z^8 = r^8 e^{i8\theta}\\ z^8 + \bar{z}^8 = r^8(e^{i8\theta}+e^{-i8\theta}) = 2r^8\cos(8\theta)\\ z = \dfrac{3+i\sqrt{3}}{2}\\ r^2 = \left(\dfrac 3 2\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 = 3\\ r^8 = \left(r^2\right)^4 = 81\\ \theta = \arctan\left(\dfrac{\frac{\sqrt{3}}{2}}{\frac 3 2}\right) = \\ \arctan\left(\dfrac{\sqrt{3}}{3}\right) = \dfrac \pi 6\\ 8\theta = \dfrac{4\pi}{3}\\ \cos(8\theta) = -\dfrac 1 2\\ 2r^8 \cos(8\theta) = 2 \cdot 81 \cdot -\dfrac 1 2= -81\)

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 Aug 18, 2019
edited by Rom  Aug 18, 2019
 #3
avatar+107018 
+1

Thanks Heureka and Rom.

 

You both reminded me of things I had almost forgotten.  Thanks :)

 Aug 18, 2019
 #4
avatar+23884 
+2

Thank you, Melody !

 

laugh

heureka  Aug 19, 2019

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