(A) Given expression: $(-1+i \sqrt{3})^{60}$ We can write this as: $2^{60} \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right)^{60}$ Let $\omega = -\fr

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(A) Given expression: $(-1+i \sqrt{3})^{60}$ We can write this as: $2^{60} \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right)^{60}$ Let $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$ be the cube root of unity. Then the expression becomes: $2^{60} \times \omega^{60}$ Since $\omega^3 = 1$,we have $\omega^{60} = (\omega^3)^{20} = 1^{20} = 1$. Therefore,$(-1+i \sqrt{3})^{60} = 2^{60} \times 1 = 2^{60}$.

Class 11 Mathematics 4-1.Complex numbers De Moivre's theorem and Roots of unity

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$(-1+i \sqrt{3})^{60} = ?$

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A
$2^{60}$
B
$2^{59}$
C
$2^{61}$
D
$2^{30}$

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$(-1+i \sqrt{3})^{60} = ?$

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