What is the value of (1-i sqrt{3})^9? | vedclass

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(B) Given expression: $(1-i \sqrt{3})^9$We can write this as: $2^9 \left(\frac{1-i \sqrt{3}}{2}\right)^9$$= 2^9 \left(\frac{1}{2} - i \frac{\sqrt{3}}{

Vedclass · Accepted Answer

$-2^9$

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(B) Given expression: $(1-i \sqrt{3})^9$We can write this as: $2^9 \left(\frac{1-i \sqrt{3}}{2}ight)^9$$= 2^9 \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}ight)^9$Using polar form: $\frac{1}{2} - i \frac{\sqrt{3}}{2} = \cos\left(-\frac{\pi}{3}ight) + i \sin\left(-\frac{\pi}{3}ight)$So,$2^9 \left[\cos\left(-\frac{\pi}{3}ight) + i \sin\left(-\frac{\pi}{3}ight)ight]^9$Applying De Moivre's theorem,$(\cos 	heta + i \sin 	heta)^n = \cos(n	heta) + i \sin(n	heta)$:$= 2^9 [\cos(-3\pi) + i \sin(-3\pi)]$$= 2^9 [\cos(3\pi) - i \sin(3\pi)]$Since $\cos(3\pi) = -1$ and $\sin(3\pi) = 0$:$= 2^9 [-1 - 0] = -2^9$

What is the value of $(1-i \sqrt{3})^9$?

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