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If z is a complex number satisfying z + 1/z =1, calculate z^10+ 1/z^10. 
 

 Jun 26, 2020
 #1
avatar+25490 
+3

If

\(z\) is a complex number satisfying \(z + \dfrac{1}{z} =1\), calculate \(z^{10}+ \dfrac{1}{z^{10}}\)

 

\(\begin{array}{|rcll|} \hline \left( z+\dfrac{1}{z} \right)^2 &=& z^2+\dfrac{1}{z^2} + \dfrac{2z}{z} \\\\ 1^2 &=& z^2+\dfrac{1}{z^2} + 2 \\\\ z^2+\dfrac{1}{z^2} &=& 1-2 \\\\ \mathbf{z^2+\dfrac{1}{z^2}} &=& \mathbf{-1} \\ \hline \end{array} \begin{array}{|rcll|} \hline \left( z^2+\dfrac{1}{z^2} \right)\left( z+\dfrac{1}{z} \right) &=& z^3+\dfrac{1}{z^3} + z+\dfrac{1}{z} \\\\ (-1)*1 &=& z^3+\dfrac{1}{z^3} + 1 \\\\ z^3+\dfrac{1}{z^3} &=&-1-1 \\\\ \mathbf{z^3+\dfrac{1}{z^3}} &=& \mathbf{-2} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \left( z^2+\dfrac{1}{z^2} \right)\left( z^3+\dfrac{1}{z^3} \right) &=& z^5+\dfrac{1}{z^5} + z+\dfrac{1}{z} \\\\ (-1)*(-2) &=& z^5+\dfrac{1}{z^5} + 1 \\\\ z^5+\dfrac{1}{z^5} &=&2-1 \\\\ \mathbf{z^5+\dfrac{1}{z^5}} &=& \mathbf{1} \\ \hline \end{array} \begin{array}{|rcll|} \hline \left( z^5+\dfrac{1}{z^5} \right)^2 &=& z^{10}+\dfrac{1}{z^{10}} +\dfrac{2z^5}{z^5} \\\\ 1^2 &=& z^{10}+\dfrac{1}{z^{10}} + 2 \\\\ z^{10}+\dfrac{1}{z^{10}} &=&1-2 \\\\ \mathbf{z^{10}+\dfrac{1}{z^{10}}} &=& \mathbf{-1} \\ \hline \end{array}\)

 

edited: Thank you Alan, thank you MaxWong

 

also see: https://web2.0calc.com/questions/if-z-is-a-complete-number-satisfying

 

laugh

 Jun 26, 2020
edited by heureka  Jun 26, 2020
edited by heureka  Jun 29, 2020
edited by heureka  Jun 29, 2020
edited by heureka  Jun 29, 2020
 #3
avatar+8336 
+5

heureka:

 

\(\left(z^2 + \dfrac1{z^2}\right)\left(z^5 + \dfrac1{z^5}\right) = z^{\color{red}7\color{black}} + \dfrac1{z^{\color{red}7\color{black}}} + z^3 + \dfrac1{z^3}\)

MaxWong  Jun 26, 2020
 #5
avatar+25490 
+4

Thank you, MaxWong

 

laugh

heureka  Jun 29, 2020
 #2
avatar+8336 
+3

An interesting theorem: If \(z + \dfrac1z = 2\cos \alpha\), then \(z^n + \dfrac1{z^n} = 2\cos n\alpha\) for positive integers n and real \(\alpha\).

 

(If you don't like to read proofs, then you can jump to the point where I marked "END OF PROOF".)

 

Proof of the theorem:

We use strong induction.

 

Base case #1 : n = 1.

Substituting n = 1 and checking shows that base case #1 is true.

 

Base case #2 : n = 2.

\(z^2 + \dfrac1{z^2} = \left(z + \dfrac1z\right)^2 - 2 = (2\cos \alpha)^2 - 2=2(2\cos^2 \alpha - 1) = 2\cos 2\alpha\)

So base case #2 is true.

 

Inductive step: Assume \(z^t + \dfrac1{z^t} = 2\cos t\alpha\) and \(z^{t - 1} + \dfrac1{z^{t - 1}} = 2\cos(t -1)\alpha\) for some positive integer \(t\).

\(\quad z^{t + 1} + \dfrac1{z^{t + 1}}\\ = \left(z^t + \dfrac1{z^t}\right)\left(z + \dfrac1z\right) - \left(z^{t - 1} + \dfrac1{z^{t - 1}}\right)\\ = 2\cos t\alpha \cdot 2\cos \alpha - 2\cos (t - 1)\alpha\\ \text{By product to sum formula:}\\ = 2 \left(\cos (t + 1)\alpha + \cos(t - 1)\alpha\right) - 2\cos(t-1)\alpha\\ =2\cos(t + 1)\alpha\)

Therefore by principles of mathematical induction, the proposition is true.

 

-------------------------------------------------------------------------(END OF PROOF)-------------------------------------------------------------------------

 

Back to the problem. Using the theorem I mentioned above, and substituting \(\alpha = 60^\circ\) into the formula,

 

\(\quad z^{10} + \dfrac1{z^{10}}\\ = 2\cos\left(10\cdot 60^\circ\right)\\ = 2 \cos\left(600^\circ\right)\\ = 2 \cos 240^\circ\\ = \boxed{-1}\)

 Jun 26, 2020
edited by MaxWong  Jun 26, 2020
 #4
avatar+43 
+7

Thank you so much! 

 Jun 27, 2020

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