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I was doing the following question:

Show $(1+\sqrt{3}i)^9 + (1-\sqrt{3}i)^9 + 2^{10} = 0$

Hint show $(1+\sqrt{3}i)^9 = (1-\sqrt{3}i)^9 = (-2)^9$

I got $1+\sqrt{3}i = 1-\sqrt{3}i = 2(cos{\frac{\pi}{3}} + i \cdot \sin{\frac{\pi}{3}})$

Then used De Moivre's and got

$2(cos{\frac{9\pi}{3}} + i \cdot \sin{\frac{9\pi}{3}})$

$= 2(cos{3\pi} + i \cdot \sin{3\pi})$

$= 2(cos{\pi} + i \cdot \sin{\pi})$

$= -2$

I missed the hint saying I should leave $x^9$ for later. So question is it appears that I can have 2 different answers, 1 if I use De Moivre 1st one don't use?

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    Oh yes, my typo there - corrected2011-11-23

1 Answers 1

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$\frac{1+\sqrt{3}i}{2}=\cos(\pi/3)+i\sin(\pi/3)=e^{i\pi/3}$

$\frac{1-\sqrt{3}i}{2}=\cos(\pi/3)-i\sin(\pi/3)=\cos(-\pi/3)+i\sin(-\pi/3)=e^{-i\pi/3}$

$(1+\sqrt{3}i)^9=2^9(e^{i\pi/3})^9$

$(1-\sqrt{3}i)^9=2^9(e^{-i\pi/3})^9$

$(1+\sqrt{3}i)^9+(1-\sqrt{3}i)^9=2^9(e^{i\pi/3})^9+2^9(e^{-i\pi/3})^9$

$(1+\sqrt{3}i)^9+(1-\sqrt{3}i)^9=2^9e^{3\pi i}+2^9e^{-3\pi i}=-2^9-2^9=-2^{10}$