If i = sqrt{-1},then 4 + 5left(-frac{1}{2} + | vedclass

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(C) Let $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ be the complex cube root of unity,where $\omega^3 = 1$.Given expression is $4 + 5\omega^{334} +

Vedclass · Accepted Answer

$i\sqrt{3}$

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(C) Let $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ be the complex cube root of unity,where $\omega^3 = 1$.Given expression is $4 + 5\omega^{334} + 3\omega^{365}$.Since $\omega^3 = 1$,we have $\omega^{334} = \omega^{333} \cdot \omega = (\omega^3)^{111} \cdot \omega = 1^{111} \cdot \omega = \omega$.Similarly,$\omega^{365} = \omega^{363} \cdot \omega^2 = (\omega^3)^{121} \cdot \omega^2 = 1^{121} \cdot \omega^2 = \omega^2$.Substituting these values into the expression:$4 + 5\omega + 3\omega^2$.Using the property $1 + \omega + \omega^2 = 0$,we have $\omega^2 = -1 - \omega$.Substituting this:$4 + 5\omega + 3(-1 - \omega) = 4 + 5\omega - 3 - 3\omega = 1 + 2\omega$.Substituting $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$:$1 + 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 1 - 1 + i\sqrt{3} = i\sqrt{3}$.

If $i = \sqrt{-1}$,then $4 + 5\left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)^{334} + 3\left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)^{365}$ is equal to

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