(1+sqrt{3}i)^6-(sqrt{3}+i)^6= | vedclass

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(D) Let $z_1 = 1+\sqrt{3}i$. We can write this in polar form as $z_1 = 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$. By De Moivre's Theorem,$z_1^6 =

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(D) Let $z_1 = 1+\sqrt{3}i$. We can write this in polar form as $z_1 = 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$. By De Moivre's Theorem,$z_1^6 = 2^6(\cos(6 	imes \frac{\pi}{3}) + i\sin(6 	imes \frac{\pi}{3})) = 64(\cos(2\pi) + i\sin(2\pi)) = 64(1+0) = 64$. Let $z_2 = \sqrt{3}+i$. We can write this in polar form as $z_2 = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$. By De Moivre's Theorem,$z_2^6 = 2^6(\cos(6 	imes \frac{\pi}{6}) + i\sin(6 	imes \frac{\pi}{6})) = 64(\cos(\pi) + i\sin(\pi)) = 64(-1+0) = -64$. Therefore,$(1+\sqrt{3}i)^6-(\sqrt{3}+i)^6 = 64 - (-64) = 64 + 64 = 128$.

$(1+\sqrt{3}i)^6-(\sqrt{3}+i)^6=$

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