ISI2018-MMA-9

Question

If $\alpha$ is a root of $x^2-x+1$, then $\alpha^{2018} + \alpha^{-2018}$ is

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

$x^2-x+1=0$

$\Rightarrow x=\frac{1\pm\sqrt{3}i}{2}=-\omega,-\omega^2$

(Where $\omega$ is the cube root of unity)$

Product of roots $\alpha \beta=1$

$\implies \beta=\alpha^{-1}$

$\alpha^{2018}+\alpha^{-2018}=\alpha^{2018}+\beta^{2018}$

$=(-\omega)^{2018}+(-\omega^2)^{2018}$

$=(\omega^3)^{672}.\omega^2+(\omega^3)^{2*672}.\omega^4$

$=\omega^2+\omega^4$

$=\omega^2+ \omega$

$=-1$      (option A)

($\color{blue}{1+\omega+\omega^2=0\;  and \; \omega^3=1}$)

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Cube root of unity--

$x^3=1$

$\Rightarrow x^3-1=0$

$\Rightarrow (x-1)(x^2+x+1)=0$

$\Rightarrow x=1,\frac{-1\mp\sqrt 3 i}{2}$