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Let \(z = \frac{\sqrt{2 + \sqrt{3}}}{2} + i \cdot \frac{\sqrt{2 - \sqrt{3}}}{2}\)

 

Find \(1 + z + z^2 + z^3 + \dots + z^{2016}\)

 May 22, 2020
 #1
avatar+30028 
+3

The sum is 1.  See the following:

 

 
 May 22, 2020
 #2
avatar+24949 
+2

Let
\(z = \frac{\sqrt{2 + \sqrt{3}}}{2} + i \cdot \frac{\sqrt{2 - \sqrt{3}}}{2}\)
Find

\(1 + z + z^2 + z^3 + \dots + z^{2016}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{z} &=& \mathbf{\frac{\sqrt{2 + \sqrt{3}}}{2} + i \cdot \frac{\sqrt{2 - \sqrt{3}}}{2}} \\\\ z &=& \sqrt{ \left( \frac{\sqrt{2 + \sqrt{3}}}{2} \right)^2 + \left( \frac{\sqrt{2 - \sqrt{3}}}{2} \right)^2 } e^{ i* \arctan\left( \frac{\sqrt{2 - \sqrt{3}}} {\sqrt{2 + \sqrt{3}}} \right) } \\\\ z &=& 1* e^{ i*\frac{\pi}{12} } \\\\ \mathbf{z} &=& \mathbf{e^{ i*\frac{\pi}{12} }} \\ \hline \end{array} \)

 

Sum of a geometric sequence

\(\begin{array}{|rcll|} \hline \mathbf{1 + z + z^2 + z^3 + \dots + z^{2016}} &=& \mathbf{\dfrac{z^{2017}-1}{z-1}} \\\\ &=& \dfrac{\left( e^{ i*\frac{\pi}{12} } \right)^{2017}-1}{e^{ i*\frac{\pi}{12} }-1} \\\\ &=& \dfrac{ e^{ i*\frac{2017}{12}\pi } -1}{e^{ i*\frac{\pi}{12} }-1} \\\\ &=& \dfrac{ e^{ i*30255^\circ } -1}{e^{ i*15^\circ }-1} \\\\ &=& \dfrac{ \cos(30255^\circ) +i\sin(30255^\circ) -1}{\cos(15^\circ) +i\sin(15^\circ)-1} \\\\ &=& \dfrac{ \cos(15^\circ) +i\sin(15^\circ) -1}{\cos(15^\circ) +i\sin(15^\circ)-1} \\\\ \mathbf{1 + z + z^2 + z^3 + \dots + z^{2016}} &=& \mathbf{1} \\ \hline \end{array}\)

 

laugh

 
 May 22, 2020

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