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Let z = 2 e^{(10 \pi i)/21}  and w = e^{(\pi i)/7}.Then what is |(z+w)^6|, the magnitude of (z+w)^6?

 May 6, 2019
 #1
avatar+27837 
+4

Like so:

 

 May 6, 2019
 #2
avatar+22180 
+2

Let \(z = 2 e^{(10 \pi i)/21}\)  and \(w = e^{(\pi i)/7}\).

Then what is \(|(z+w)^6|\),

the magnitude of \((z+w)^6\)?

 

\(\begin{array}{|rcll|} \hline z = 2 e^{i\left(\frac{10}{21} \pi \right) } && w = e^{i\left(\frac{\pi}{7} \right) } \\ \boxed{\frac{10}{21} \pi = \frac{7}{21}\pi + \frac{3}{21}\pi \\ = \frac{1}{3}\pi + \frac{1}{7}\pi } \\ z = 2 e^{i\left(\frac{1}{3}\pi + \frac{1}{7}\pi \right) } \\ z = 2 e^{i\left(\frac{\pi}{3} \right) }e^{i\left(\frac{\pi}{7} \right) } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline z+w &=& 2 e^{i\left(\frac{\pi}{3} \right) }e^{i\left(\frac{\pi}{7} \right) } + e^{i\left(\frac{\pi}{7} \right) } \\ &=& e^{i\left(\frac{\pi}{7} \right)} \left( 2 e^{i\left(\frac{\pi}{3} \right)}+1 \right) \\ && \boxed{ e^{i\left(\frac{\pi}{3} \right)} = \cos\left(\frac{\pi}{3} \right) + i\sin\left(\frac{\pi}{3} \right) \\ = \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \\ 2e^{i\left(\frac{\pi}{3} \right)} =1+i\sqrt{3} } \\\\ &=& e^{i\left(\frac{\pi}{7} \right)} \left(1+i\sqrt{3} +1 \right) \\ &=& e^{i\left(\frac{\pi}{7} \right)} \left(2+i\sqrt{3} \right) \\\\ && \boxed{2+i\sqrt{3} = \sqrt{2^2+(\sqrt{3})^2} \cdot e^{i\cdot \arctan(\frac{\sqrt{3}}{2} ) } \\ = \sqrt{7} \cdot e^{i\cdot \arctan(\frac{\sqrt{3}}{2} ) } } \\\\ &=& \sqrt{7} \cdot e^{i\cdot \arctan(\frac{\sqrt{3}}{2} ) } e^{i\left(\frac{\pi}{7} \right)} \\ &=& \sqrt{7} \cdot e^{i\cdot \arctan(\frac{\sqrt{3}}{2} ) +i\left(\frac{\pi}{7} \right)} \\ &=& \sqrt{7} \cdot e^{i\cdot \arctan(\frac{\sqrt{3}}{2} + \frac{\pi}{7} ) } \\ \left(z+w \right)^6 &=& \left(\sqrt{7}\right)^6 \cdot e^{i\cdot 6\arctan(\frac{\sqrt{3}}{2} + \frac{\pi}{7} ) } \\ |\left(z+w \right)^6| &=& \left(\sqrt{7}\right)^6 \\ &=& 7^3 \\ \mathbf{|\left(z+w \right)^6|} &=& \mathbf{343} \\ \hline \end{array}\)

 

...Thank you, Alan...

 

laugh

 May 6, 2019
edited by heureka  May 6, 2019
edited by heureka  May 6, 2019
edited by heureka  May 7, 2019

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