## Task B-4.4.

If $z$ is a solution to the equation $z^{2}-z+1=0$, what is $z^{2010}-z^{1005}+1$?
## First solution.

The solution to the given equation is not $z=-1$, so we can multiply that equation by $z+1$. We will get:

$$
\begin{aligned}
(z+1)\left(z^{2}-z+1\right) & =0 \\
z^{3}+1 & =0 \\
z^{3} & =-1
\end{aligned}
$$

Then $z^{1005}=\left(z^{3}\right)^{335}=-1$,

and $z^{2010}=\left(z^{1005}\right)^{2}=1$

so $z^{2010}-z^{1005}+1=3$.