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  4. ((-1+√-3/2))100 + ((-1-√-3/2))100 is equal to

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Q. $\left(\frac{-1+\sqrt{-3}}{2}\right)^{100} + \left(\frac{-1-\sqrt{-3}}{2}\right)^{100}$ is equal to

Complex Numbers and Quadratic Equations

Solution:

Let $\frac{-1+i\sqrt{-3}}{2} = \omega$ then $\frac{-1-\sqrt{3}i}{2} = \omega^{2}$
where $\omega =$ cube root of unity
Consider $\left(\frac{-1+\sqrt{-3}}{2}\right)^{100} + \left(\frac{-1-\sqrt{-3}}{2}\right)^{100}$
$= \left(\frac{-1+\sqrt{3}i}{2}\right)^{100} + \left(\frac{-1-\sqrt{3}i}{2}\right)^{100}$
$= \omega^{100} + \omega^{200} = \omega+\omega^{2} = -1$
($\therefore \omega^{3} = 1$ and $1 + \omega+\omega^{2} = 0$)