Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$. 
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:


*

*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$

*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ and $z_2=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}$

*How do I proceed?

 A: Using $e^{i x} = \cos(x) + i \sin(x)$ then for $z_{1} = e^{2 \pi i/3}$ it is seen that 
\begin{align}
z_{1}^{10} + \frac{1}{z_{1}^{10}} &= e^{20 \pi i/3} + e^{- 20 \pi i/3} = e^{6 \pi i + 2 \pi i/3} + e^{-6 \pi i - 2\pi i/3} \\
&= e^{2\pi i/3} + e^{-2\pi i/3} = 2 \cos(2\pi /3) = -1 
\end{align}
A: Hints:


*

*Note $z_1z_2=1$

*You could write your (2) as $z_1=e^{i2\pi/3}$ and $z_2=e^{-i2\pi/3}$. 

*Taking the $10$th power of an exponential is easy
A: Hint: $z^2+z+1=0$ implies that $0=(z-1)(z^2+z+1)=z^3-1$, so $z^3=1$. Also $z^{10}=z^9\cdot z$ and $z^{-10}=z^2\cdot z^{-12}$.
A: The solutions to $z^2 + z + 1 = 0$ are the complex $z = 1 \angle (\pm 1/3)$.
So 
$$\begin{align}
z^{10} + z^{-10} & = \left(1\angle (\pm 1/3)\right)^{10} + \left(1\angle (\pm 1/3)\right)^{-10} 
\\& = \left(1\angle (\pm 10/3)\right) + \left(1\angle (\mp 10/3)\right) \
\\ & = (1\angle 10/3) + (1\angle -10/3)
\\ & = (1\angle 1/3) + (1\angle -1/3)
\\ & = -1
\end{align}$$
A: $z^2+z+1 = 0 \implies   z^2+z = -1 $
$z^2= -z-1 \implies z^3 = -z^2-z = -(z^2+z)= 1$
$z^{10} = z^3\cdot z^3\cdot z^3\cdot z=  z \implies$
$z^{10} + z^{-10} = z + 1/z = \frac{z^2 +1 }z$ but $z^2+1 = -z$ so  this results in $-z/z = -1$
