The imaginary part of (sqrt{3}-i)^{2016}+(-sq | vedclass

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(C) Let $z_1 = \sqrt{3}-i = 2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6}))$. Then $z_1^{2016} = 2^{2016}(\cos(-\frac{2016\pi}{6}) + i\sin(-\frac{2016

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$-2^{2019}$

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(C) Let $z_1 = \sqrt{3}-i = 2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6}))$. Then $z_1^{2016} = 2^{2016}(\cos(-\frac{2016\pi}{6}) + i\sin(-\frac{2016\pi}{6})) = 2^{2016}(\cos(-336\pi) + i\sin(-336\pi)) = 2^{2016}(1 + 0i) = 2^{2016}$.Let $z_2 = -\sqrt{3}-i = 2(\cos(-\frac{5\pi}{6}) + i\sin(-\frac{5\pi}{6}))$. Then $z_2^{2019} = 2^{2019}(\cos(-\frac{2019 	imes 5\pi}{6}) + i\sin(-\frac{2019 	imes 5\pi}{6})) = 2^{2019}(\cos(-\frac{3365\pi}{2}) + i\sin(-\frac{3365\pi}{2}))$.Since $-\frac{3365\pi}{2} = -1682\pi - \frac{\pi}{2}$,we have $\cos(-\frac{3365\pi}{2}) = \cos(-\frac{\pi}{2}) = 0$ and $\sin(-\frac{3365\pi}{2}) = \sin(-\frac{\pi}{2}) = -1$.Thus,$z_2^{2019} = 2^{2019}(0 - i) = -i 2^{2019}$.The sum is $2^{2016} - i 2^{2019}$.The imaginary part is $-2^{2019}$.

The imaginary part of $(\sqrt{3}-i)^{2016}+(-\sqrt{3}-i)^{2019}$ is

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