$(z + 1/z)^2 + (z^2 + 1/z^2)^2 + (z^3 + 1/z^3)^2 + (z^4 + 1/z^4)^4 + (z^5 + 1/z^5)^2 + (z^6 + 1/z^6)^2$ if $z^2 + z + 1 = 0$

Question

Solve $(z + \frac{1}{z})^2 + (z^2 + \frac{1}{z^2})^2 + (z^3 + \frac{1}{z^3})^2 + (z^4 + \frac{1}{z^4})^2 + (z^5 +\frac{1}{z^5})^2 + (z^6 + \frac{1}{z^6})^2$

if $z^2 + z + 1 = 0$

I tried this problem and come up with this solution:

$z^2 + z + 1 = 0$

$z^2 = -z -1$

$(z^2)^2 = (-z -1)^2$

$z^4 + (z^2 + 1) = z^2+2z+1+ (z^2 + 1)$

$z^4 + z^2 + 1 = 2(z^2 + z + 1) = 0$

and this help me to figure out that the answer is 12. However the step $(z^2)^2 = (-z -1)^2$ isn't quite valid because it turns a quadratic equation to a quartic equation, which will lead to extreneous roots.

Therefore, I want to ask for a better solution to this. Thanks in advance.

Answer 1

$z+\dfrac1z=-1$

$z^2+\dfrac1{z^2}=\left(z+\dfrac1z\right)^2-2=?$

$z^3-1=(z-1)(z^2+z+1)=?$

Answer 2

If lab's $z^3=1$ idea doesn't strike you immediately here is a dumb method for working sum of powers of roots. $$t^2+t+1=0$$ Notice that if you replace $t$ by $\frac{1}{t}$, the equation doesn't change.
This means if $z$ is a root, then $\frac{1}{z}$ is also a root.

Using sum of roots, product of roots:
$s_1=z+\frac{1}{z} = -1$
$s_2=z^2+\frac{1}{z^2} = s_1^2 - 2 = -1$

Next use the following recurrence to compute remaining sum of powers $$s_{n+1}=-s_n-s_{n-1}$$