+0  
 
0
40
3
avatar

if z is a complex number such that \(z+1/z= \sqrt{3}\)

 

what is the value of \(z^{2010}+1/z^{2010}\)?

 

Thanks for any hints, help, or answers!

 May 31, 2021
 #1
avatar
0

By Demoivre's Theorem, z^2010 + 1/z^2010 = 1.

 May 31, 2021
 #2
avatar
0

After answering 1, it said it was incorrect and that the actual answer is -2.  Thanks for answering though!

Guest May 31, 2021
 #3
avatar+25928 
+2

if z is a complex number such that \(z+\dfrac{1}{z}= \sqrt{3}\)
what is the value of \(z^{2010}+\dfrac{1}{z^{2010}}\)?

 

1. \(|z| = ~?\)

\(\begin{array}{|rcll|} \hline z+\dfrac{1}{z}&=& \sqrt{3} \quad | \quad \text{Let $z=a+bi$} \\\\ a+bi+\dfrac{1}{a+bi}&=& \sqrt{3} \\\\ a+bi+\dfrac{1}{(a+bi)}*\dfrac{(a-bi)}{(a-bi)}&=& \sqrt{3}\quad | \quad (a+bi)*(a-bi)=|z|^2 \\\\ a+bi+\dfrac{a-bi}{|z|^2}&=&\sqrt{3} \\\\ \underbrace{ a+\dfrac{a}{|z|^2}}_{=\sqrt{3}}+ \underbrace{ bi-\dfrac{bi}{|z|^2} }_{=0}&=&\sqrt{3} \\\\ \hline bi-\dfrac{bi}{|z|^2} &=& 0 \\\\ bi &=& \dfrac{bi}{|z|^2} \\\\ |z|^2 &=& \dfrac{bi}{bi} \\\\ \mathbf{ |z|^2 } &=& \mathbf{1} \quad \text{or} \quad \mathbf{|z|=1} \\ \hline \end{array}\)

 

2. \(\varphi= ~?\)

\(\begin{array}{|rcll|} \hline z&=& |z|*e^{i\varphi} \quad | \quad |z|=1 \\ z&=& e^{i\varphi} \\ \mathbf{z}&=& \mathbf{\cos(\varphi)+i*\sin(\varphi)} \\ \hline \dfrac{1}{z}&=&z^{-1} \\ &=& \left( |z|*e^{i\varphi} \right)^{-1} \\ &=& |z|^{-1}* \left(e^{i\varphi}\right)^{-1} \\ &=& \dfrac{1}{|z|}*e^{i\varphi(-1)} \\ &=& \dfrac{1}{|z|}*e^{i(-\varphi} \quad | \quad |z|=1 \\ &=& e^{i(-\varphi} \\ &=& \cos(-\varphi)+i*\sin(-\varphi) \\ \mathbf{\dfrac{1}{z}} &=& \mathbf{\cos(\varphi)-i*\sin(\varphi)} \\ \hline z+\dfrac{1}{z} &=& \cos(\varphi)+i*\sin(\varphi)+\cos(\varphi)-i*\sin(\varphi) \\\\ z+\dfrac{1}{z} &=& 2\cos(\varphi) \quad | \quad z+\dfrac{1}{z} = \sqrt{3} \\ \sqrt{3} &=& 2\cos(\varphi) \\ \cos(\varphi) &=& \dfrac{\sqrt{3}}{2} \\\\ \mathbf{\varphi} &=& \mathbf{30^\circ} \\ \hline \end{array}\)

 

3. \(z^{2010}=~?\)

\(\begin{array}{|rcll|} \hline z^{2010} &=& \left( |z|*e^{i\varphi} \right)^{2010} \\ &=& |z|^{2010}* \left(e^{i\varphi}\right)^{2010} \\ &=& 1^{2010}*e^{i*2010\varphi} \\ &=& e^{i*2010\varphi} \\ &=& \cos(2010\varphi)+i*\sin(2010\varphi) \quad | \quad \varphi=30^\circ\\ &=& \cos(2010*30^\circ)+i*\sin(2010*30^\circ)\\ &=& \cos(180^\circ)+i*\sin(180^\circ)\\ \mathbf{z^{2010}}&=& \mathbf{-1} \\ \hline \end{array}\)

 

4. \(\dfrac{1}{z^{2010}}=~?\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{z^{2010}} &=& \left( |z|*e^{i\varphi} \right)^{-2010} \\ &=& |z|^{-2010}* \left(e^{i\varphi}\right)^{-2010} \\ &=& 1^{-2010}* e^{i*(-2010\varphi)} \\ &=& e^{i*(-2010\varphi)} \\ &=& \cos(-2010\varphi)+i*\sin(-2010\varphi) \\ &=& \cos(2010\varphi)-i*\sin(2010\varphi) \quad | \quad \varphi=30^\circ\\ &=& \cos(2010*30^\circ)-i*\sin(2010*30^\circ) \\ &=& \cos(180^\circ)-i*\sin(180^\circ) \\ \mathbf{\dfrac{1}{z^{2010}}}&=& \mathbf{-1} \\ \hline \end{array}\)

 

\(z^{2010} + \dfrac{1}{z^{2010}} = -1+(-1) \\ z^{2010} + \dfrac{1}{z^{2010}} = -2 \)

 

laugh

 Jun 1, 2021

40 Online Users

avatar
avatar
avatar
avatar
avatar