1. If $z=\frac{1+i}{\sqrt{2}}$, calculate the sum

$$
1+z+z^{2}+z^{3}+\ldots+z^{2006}
$$
Solution.

Notice that

$$
\begin{aligned}
& z=\frac{1+i}{\sqrt{2}}, \quad z^{2}=i, \quad z^{3}=\frac{-1+i}{\sqrt{2}}, \quad z^{4}=-1 \\
& z^{5}=\frac{-1-i}{\sqrt{2}}, \quad z^{6}=-i, \quad z^{7}=\frac{1-i}{\sqrt{2}}, \quad z^{8}=1
\end{aligned}
$$

so we have

$$
1+z+z^{2}+\ldots+z^{7}=0
$$

Since $z^{k}=z^{8 l+k}, k \in\{0,1,2, \ldots, 7\}, l \in \mathbb{N}$, and since $2006=8 \cdot 250+6$, the given sum $S$ can be written in the form

$$
\begin{aligned}
S= & \left(1+z+z^{2}+\ldots+z^{7}\right)\left(1+z^{8}+z^{2 \cdot 8}+\ldots+z^{249 \cdot 8}\right) \\
& +\left(1+z+z^{2}+\ldots+z^{6}\right) \cdot z^{250 \cdot 8} \\
= & 1+z+z^{2}+\ldots+z^{6}=\frac{-1+i}{\sqrt{2}}
\end{aligned}
$$