Given z = (1 + isqrt{3})^{100},then frac{text | vedclass

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(C) Let $w = 1 + i\sqrt{3}$.Converting to polar form,$r = |w| = \sqrt{1^2 + (\sqrt{3})^2} = 2$.$	heta = \arg(w) = 	an^{-1}(\sqrt{3}/1) = \frac{\pi}{

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$\frac{1}{\sqrt{3}}$

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(C) Let $w = 1 + i\sqrt{3}$.Converting to polar form,$r = |w| = \sqrt{1^2 + (\sqrt{3})^2} = 2$.$	heta = \arg(w) = 	an^{-1}(\sqrt{3}/1) = \frac{\pi}{3}$.So,$w = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})$.Then $z = w^{100} = 2^{100}(\cos \frac{100\pi}{3} + i \sin \frac{100\pi}{3})$.Since $\frac{100\pi}{3} = 33\pi + \frac{\pi}{3}$,we have $\cos(\frac{100\pi}{3}) = \cos(33\pi + \frac{\pi}{3}) = -\cos(\frac{\pi}{3}) = -\frac{1}{2}$ and $\sin(\frac{100\pi}{3}) = \sin(33\pi + \frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$.Thus,$z = 2^{100}(-\frac{1}{2} - i\frac{\sqrt{3}}{2})$.$	ext{Re}(z) = -2^{99}$ and $	ext{Im}(z) = -2^{99}\sqrt{3}$.Therefore,$\frac{	ext{Re}(z)}{	ext{Im}(z)} = \frac{-2^{99}}{-2^{99}\sqrt{3}} = \frac{1}{\sqrt{3}}$.

Given $z = (1 + i\sqrt{3})^{100}$,then $\frac{\text{Re}(z)}{\text{Im}(z)}$ equals

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