Biết rằng số phức $Z$ thoả mãn $Z+\frac{1}{Z}=1$ hãy tính $Z^{2010}+\frac{1}{Z^{2010}}$.
Từ $Z+\frac{1}{Z}=1\Leftrightarrow Z^2-Z+1=0\Leftrightarrow \left[ \begin{array}{l}Z=\frac{1}{2}+\frac{\sqrt{3}}{2}i=\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}  \\ Z=\frac{1}{2}-\frac{\sqrt{3}}{2}i=\cos (-\frac{\pi}{3})+i\sin (-\frac{\pi}{3})\end{array} \right.$
a) $Z=\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}\Rightarrow Z^{2010}=\cos \frac{2010\pi}{3}+i\sin \frac{2010\pi}{3}  (1)$
Ta có: $\frac{1}{Z}=\frac{1}{\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}}=\cos \frac{\pi}{3}-i\sin \frac{\pi}{3}=\cos (-\frac{\pi}{3})+i\sin (-\frac{\pi}{3})$
$\Rightarrow \frac{1}{Z^{2010}}=\cos (-\frac{2010\pi}{3})+i\sin (-\frac{2010\pi}{3})   (2)$
Từ $(1),(2)$ suy ra $Z^{2010}+\frac{1}{Z^{2010}}=2\cos (670\pi)=2$.
b) Nếu $Z=\cos (-\frac{\pi}{3})+i\sin (-\frac{\pi}{3})\Rightarrow \frac{1}{Z}=\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}\Rightarrow Z^{2010}+\frac{1}{Z^{2010}}=2$.
Tóm lại ta luôn có: $Z^{2010}+\frac{1}{Z^{2010}}=2$.
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