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?