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Let \(z\) be a complex number satisfying \(z^2 + z + 1 = 0.\) Compute \(\left( z + \frac{1}{z} \right)^2 + \left( z^2 + \frac{1}{z^2} \right)^2 + \left( z^3 + \frac{1}{z^3} \right)^2 + \dots + \left( z^{45} + \frac{1}{z^{45}} \right)^2.\)
 

 Dec 27, 2018
 #1
avatar+3576 
+3

\(\text{well.. let's brute force it and see if we get any insight}\\ z^2+z+1=0\\ z = \dfrac{-1 \pm i \sqrt{3}}{2} = e^{\pm i \frac{2\pi}{3}}\\ \text{so picking either root }\\ z^n + \dfrac{1}{z^n} = e^{i\frac{2n\pi}{3}}+e^{-i\frac{2n\pi}{3}} = 2 \cos\left(\dfrac{2n\pi}{3}\right) \)

 

\(\text{looking at }a_n = 2\cos\left(\dfrac{2n\pi}{3}\right) \text{ we find that}\\ a_n = \begin{cases} -1 &n \pmod{3} \in \{1,2\}\\ 2 &n \pmod{3} = 0 \end{cases}\)

 

\(\text{and thus the original sum becomes} \\ \sum \limits_{k=1}^{15}~(1 + 1 + 4) = 15 \cdot 6 = 90\)

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 Dec 27, 2018
 #2
avatar+809 
+1

Thank you, Rom! laugh

mathtoo  Dec 28, 2018
 #3
avatar+95171 
0

Thanks Rom :)

Another interesting one for me to inspect :))

Melody  Dec 29, 2018

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