QuestionMay 5, 2026

Find (2(cos(2pi )/(3)+isin(2pi )/(3)))^5 -16-16isqrt (3) 16+16isqrt (3) 16sqrt (3)+16i -16sqrt (3)-16i

Find (2(cos(2pi )/(3)+isin(2pi )/(3)))^5 -16-16isqrt (3) 16+16isqrt (3) 16sqrt (3)+16i -16sqrt (3)-16i
Find (2(cos(2pi )/(3)+isin(2pi )/(3)))^5 -16-16isqrt (3) 16+16isqrt (3) 16sqrt (3)+16i -16sqrt (3)-16i

Solution
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Answer

-16 - 16i\sqrt{3} Explanation 1. Express in polar form Given 2(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}), r=2, \theta=\frac{2\pi}{3}. 2. Apply De Moivre's theorem **Formula:** (r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta)) Here: r^5 = 2^5 = 32 n\theta = 5 \cdot \frac{2\pi}{3} = \frac{10\pi}{3} = \frac{10\pi}{3} - 2\pi = \frac{4\pi}{3}. 3. Evaluate trigonometric values \cos\frac{4\pi}{3} = -\frac{1}{2}, \ \sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}. 4. Multiply by magnitude 32\left(-\frac12 + i\cdot -\frac{\sqrt{3}}{2}\right) = -16 - 16i\sqrt{3}.

Explanation

1. Express in polar form <br /> Given $2(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3})$, $r=2$, $\theta=\frac{2\pi}{3}$. <br /><br />2. Apply De Moivre's theorem <br /> **Formula:** $(r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))$ <br /> Here: <br />$r^5 = 2^5 = 32$ <br />$n\theta = 5 \cdot \frac{2\pi}{3} = \frac{10\pi}{3} = \frac{10\pi}{3} - 2\pi = \frac{4\pi}{3}$. <br /><br />3. Evaluate trigonometric values <br /> $\cos\frac{4\pi}{3} = -\frac{1}{2}$, $\ \sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$. <br /><br />4. Multiply by magnitude <br /> $32\left(-\frac12 + i\cdot -\frac{\sqrt{3}}{2}\right) = -16 - 16i\sqrt{3}$.
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