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