(1-i sqrt{3})^{2025}= | vedclass

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Question

(C) Let $z = 1 - i \sqrt{3}$.First,find the polar form of $z$.The modulus $r = |z| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$.The argument $	heta

Vedclass · Accepted Answer

$-2^{2025}$

Vedclass · Answer

(C) Let $z = 1 - i \sqrt{3}$.First,find the polar form of $z$.The modulus $r = |z| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$.The argument $	heta = 	an^{-1}(\frac{-\sqrt{3}}{1}) = -\frac{\pi}{3}$.So,$z = 2(\cos(-\frac{\pi}{3}) + i \sin(-\frac{\pi}{3}))$.Using De Moivre's Theorem,$z^{2025} = 2^{2025}(\cos(2025 	imes -\frac{\pi}{3}) + i \sin(2025 	imes -\frac{\pi}{3}))$.$2025 	imes -\frac{\pi}{3} = -675\pi$.Since $-675\pi = -674\pi - \pi$,the angle is coterminal with $-\pi$.$z^{2025} = 2^{2025}(\cos(-\pi) + i \sin(-\pi)) = 2^{2025}(-1 + 0i) = -2^{2025}$.

$(1-i \sqrt{3})^{2025}=$

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