Let z=1-i√32 and i=-1.

Question

Let $z=\dfrac{1-i\sqrt3}2$ and $i=\sqrt{-1}$. Then the value of $21+\displaystyle\sum_{r=1}^{21}\left(z^r+\frac1{z^r}\right)^3$ is

Answer: 13

Accepted Answer

$z=e^{-i\pi/3}$, and $z^r+\dfrac1{z^r}=2\cos\dfrac{r\pi}3$.

$$21+\sum_{r=1}^{21}8\cos^3\frac{r\pi}3.$$

Most diametrically-paired terms cancel; the residual single term gives $-1$, so the sum is $21+8(-1)=13.$

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