(-sqrt{3} + i)^{53},where i^2 = -1,is equal t | vedclass

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Question

(C) Let $z = -\sqrt{3} + i$. First,express $z$ in polar form: $z = 2\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = 2(\cos 150^\circ + i \sin 150^\c

Vedclass · Accepted Answer

$2^{53}\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)$

Vedclass · Answer

(C) Let $z = -\sqrt{3} + i$. First,express $z$ in polar form: $z = 2\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}ight) = 2(\cos 150^\circ + i \sin 150^\circ)$. Using De Moivre's Theorem,$z^{53} = 2^{53}(\cos 150^\circ + i \sin 150^\circ)^{53}$. $z^{53} = 2^{53}(\cos(53 	imes 150^\circ) + i \sin(53 	imes 150^\circ))$. $53 	imes 150^\circ = 7950^\circ$. Dividing $7950^\circ$ by $360^\circ$: $7950 = 22 	imes 360 + 30$. So,$z^{53} = 2^{53}(\cos 30^\circ + i \sin 30^\circ)$. $z^{53} = 2^{53}\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}ight)$.

$(-\sqrt{3} + i)^{53}$,where $i^2 = -1$,is equal to:

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