Task A-2.6. (10 points)

Let $z=\frac{1}{2}+i \frac{\sqrt{3}}{2}$. Calculate

$$
z+z^{2}+z^{3}+z^{4}+\ldots+z^{k}+\ldots+z^{2010}
$$
## Solution.

We calculate: $z^{2}=-\frac{1}{2}+i \frac{\sqrt{3}}{2}, z^{3}=-1, z^{4}=-\frac{1}{2}-i \frac{\sqrt{3}}{2}, z^{5}=\frac{1}{2}-i \frac{\sqrt{3}}{2}, z^{6}=1$. (5 points) By adding, we get $z+z^{2}+z^{3}+z^{4}+z^{5}+z^{6}=0$.

Since 2010 is divisible by 6, we have

$$
\begin{aligned}
& z+z^{2}+z^{3}+z^{4}+\ldots+z^{k}+\ldots+z^{2010} \\
& =\left(z+z^{2}+\cdots+z^{6}\right)+z^{6}\left(z+z^{2}+\cdots+z^{6}\right)+\cdots+z^{2004}\left(z+z^{2}+\cdots+z^{6}\right) \\
& =\left(z+z^{2}+z^{3}+z^{4}+z^{5}+z^{6}\right)\left(1+z^{6}+z^{12}+z^{18}+\cdots+z^{2004}\right)=0
\end{aligned}
$$

Note. Using the formula for the sum of a geometric series, we immediately get:

$$
z+z^{2}+z^{3}+z^{4}+\ldots+z^{k}+\ldots+z^{2010}=z \cdot \frac{z^{2010}-1}{z-1}=z \cdot \frac{\left(z^{3}\right)^{670}-1}{z-1}=0
$$

due to the fact that $z^{3}=-1$. Such a solution should also be scored with all 10 points.