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$$z^n=r^n(\cos{n}\theta+i\sin{n}\theta)$$ $$\alpha=2+2\sqrt3\,i$$ $${\Rightarrow}\,(2+2\sqrt3\,i)^n=\left(\sqrt{2^2+(2\sqrt{3})^2}\right)^n\left(\cos\frac{n\pi}{3}+i\sin\frac{n\pi}{3}\right)$$ $${\Rightarrow}\,\alpha^n=4^n\left(\cos\frac{n\pi}{3}+i\sin\frac{n\pi}{3}\right)$$ $$\beta=2-2\sqrt3\,i$$ $${\Rightarrow}\,(2-2\sqrt3\,i)^n=\left(\sqrt{2^2+(2\sqrt{3})^2}\right)^n\left(\cos(-\frac{n\pi}{3})+i\sin(-\frac{n\pi}{3})\right)$$ $${\Rightarrow}\,\beta^n=4^n\left(\cos(-\frac{n\pi}{3})+i\sin(-\frac{n\pi}{3})\right)$$

$$\gamma=-\frac12$$ $${\Rightarrow}\,\gamma^n=\left(-\frac12\right)^n$$

$$\alpha^n+\beta^n+\gamma^n=$$ $$\left(4^n\left(\cos\frac{n\pi}{3}+i\sin\frac{n\pi}{3}\right)\right)+\left(4^n\left(\cos(-\frac{n\pi}{3})+i\sin(-\frac{n\pi}{3})\right)\right)+\left(-\frac12\right)^n$$ What have I done wrong here? The Mark Scheme has $$\,\beta^n=4^n\left(\cos(\frac{n\pi}{3})-i\sin(\frac{n\pi}{3})\right)$$

Rewriting the equation with this in mind $$\alpha^n+\beta^n+\gamma^n=$$ $$\left(4^n\left(\cos\frac{n\pi}{3}+i\sin\frac{n\pi}{3}\right)\right)+\left(4^n\left(\cos(\frac{n\pi}{3})-i\sin(\frac{n\pi}{3})\right)\right)+\left(-\frac12\right)^n$$ $${\Rightarrow}\,2\left(4^{n}\cos\frac{n\pi}{3}\right)+\left(-\frac12\right)^n$$ $${\Rightarrow}\,2\left(2^{2n}\cos\frac{n\pi}{3}\right)+\left(-\frac12\right)^n$$ Now I'm stuck, where does the $^{n+1}$ come from? And what happens to the $2$? $$2^{2n+1}\,cos\,\frac{n\pi}{3}+\left(-\frac12\right)^n$$

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    $\begingroup$ $2\cdot 2^{2n} = 2^{2n+1}$ $\endgroup$ – William Stagner Apr 5 '15 at 5:04
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If you look at the sine and cosine graphs you will see that:

$$cos{\theta}=cos(-\theta)$$ $$sin(-\theta)=-sin\theta$$ Therefore: $$\beta^n=\left(4^n\left(\cos(-\frac{n\pi}{3})+i\sin(-\frac{n\pi}{3})\right)\right)=\left(4^n\left(\cos(\frac{n\pi}{3})-i\sin(\frac{n\pi}{3})\right)\right)$$ Using the exponent rule: $$(x^m)(x^n)=x^{m+n}$$ $$2(2^{2n})=(2^1)(2^{2n})=2^{2n+1}$$

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