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The value of $\large[\frac{-1+i\sqrt{3}}{2}]^{100}+[\frac{-1-i\sqrt{3}}{2}]^{100}$ is

\[\begin{array}{1 1}(1)2&(2)0\\(3)-1&(4)1\end{array}\]

Can you answer this question?
 
 

1 Answer

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If $\omega$ is a cube root of unity then
$\omega^3=1$ and $1+\omega +\omega^2=0$
$\omega=-\large\frac{1}{2} $$ + i \large\frac{\sqrt 3}{2}$ and $\omega^2=-\large\frac{1}{2} - i \large\frac{\sqrt 3}{2}$
$\large[\frac{-1+i\sqrt{3}}{2}]^{100}+[\frac{-1-i\sqrt{3}}{2}]^{100}$
$\qquad= \omega^{100}+(\omega^{2})^{100}$
$\qquad=\omega^{100}+\omega^{200}$
$\qquad=\omega^{99}+\omega^{198}. \omega^{2}$
$\qquad=(\omega^{3})^{33}.\omega+(\omega^{3})^{66}. \omega^{2}$
$\qquad= (\omega^3)^{33}. \omega +( \omega^3)^{66}. \omega^{2}$
$\qquad=\omega+\omega^2 (\omega^3=1)$
$\qquad=-1$
Hence 3 is the correct answer.
answered May 14, 2014 by meena.p
 

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