If |x+iy|=sqrt{x^2+y^2},then |(1-sqrt{3}i)^9+ | vedclass

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(D) Let $z_1 = 1-\sqrt{3}i$ and $z_2 = \sqrt{3}+i$. Converting to polar form: $z_1 = 2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) = 2e^{-i\pi/3}$.

Vedclass · Accepted Answer

$2^{\frac{19}{2}}$

Vedclass · Answer

(D) Let $z_1 = 1-\sqrt{3}i$ and $z_2 = \sqrt{3}+i$. Converting to polar form: $z_1 = 2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) = 2e^{-i\pi/3}$. $z_2 = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})) = 2e^{i\pi/6}$. Now,$z_1^9 = 2^9 e^{-i3\pi} = 2^9(\cos(-3\pi) + i\sin(-3\pi)) = 2^9(-1) = -2^9$. $z_2^9 = 2^9 e^{i3\pi/2} = 2^9(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})) = 2^9(0 - i) = -i2^9$. Then,$|z_1^9 + z_2^9| = |-2^9 - i2^9| = |2^9(-1-i)| = 2^9|-1-i|$. $|z_1^9 + z_2^9| = 2^9 \sqrt{(-1)^2 + (-1)^2} = 2^9 \sqrt{2} = 2^9 \cdot 2^{1/2} = 2^{19/2}$. Thus,the correct option is $D$.

If $|x+iy|=\sqrt{x^2+y^2}$,then $|(1-\sqrt{3}i)^9+(\sqrt{3}+i)^9|=$

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