# Using De Moivre's Theorem Before & After

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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You must take your $2$ at the front of $2(\cos\theta+i\sin\theta)$ to the power of $9$ also. $(2(\cos\theta+i\sin\theta))^9=2^9(\cos\theta+i\sin\theta)^9$. – Jp McCarthy Nov 23 '11 at 1:34
The hint is incorrect. – picakhu Nov 23 '11 at 1:35
Something is very wrong here. It is not true that $1+\sqrt3i=1-\sqrt3i$, it is not true that $1-\sqrt3i=(-2)^9$, and it is not true that $1+\sqrt3i=(-2)^9$. – Gerry Myerson Nov 23 '11 at 1:38
The hint should probably have been $$(1+\sqrt{3}\;i)^9=(1-\sqrt{3}\;i)^9=(-2)^9$$ – robjohn Nov 23 '11 at 2:05
Oh yes, my typo there - corrected – Jiew Meng Nov 23 '11 at 2:24

## 1 Answer

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

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

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

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

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

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

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