Find $w+\frac{1}{w} + \frac{1}{w^2}$ where $w^4=1$, $w\neq 1$ Let $w$ be a root of the equation $z^4 = 1$ and $w$ is not equal to $1$. 
I would like to find the value of $w+\frac{1}{w} + \frac{1}{w^2}$.
So for this question since w is a root, 
I first substituted w and got $w^4=1$
Which is $w * w^3 = 1$. 
Since $w$ is not equal to $1$, 
$w^3=1$
Which can be rewritten as $(w-1) (1 + w + w^2) = 0$
And since $w \neq 1,$ $(1 + w + w^2) = 0$
On simplifying the given equation, 
I get 
$w^3 + w + 1
= 2+w$ (Is this right approach to go about it?)

Solution one:
The four roots are $1,-1,i,-i$. Simply plugging in the last three we get $-1$.
Solution two:
$z^4-1=0$ implies $(z^2-1)(z^2+1)=0$ implies $(z-1)(z+1)(z^2+1)=0$. Since we don't consider $1$, we may divide by $(z-1)$ to get: $(z+1)(z^2+1)=0$ or $z^3+z^2+z+1=0$.
Note $$\omega+\frac{1}{\omega}+\frac{1}{\omega^2}=\frac{\omega^3+\omega+1}{\omega^2}=\frac{-\omega^2}{\omega^2}=-1$$
