Question

If $z^2+z+1=0,z\in C$, then $\left|\sum _{n=1}^{15}{\left(Z^n+(-1{)}^n\frac{1}{Z^n}\right)}^2\right|$ is equal to_____.

Answer 1

Answer: 2

$z^2+z+1=0\Rightarrow z=w,w^2$
$\left|\sum _{n=1}^{15}{\left(z^n+(-1)\frac{1}{z^n}\right)}^2\right|=\left|\sum _{n=1}^{15}\left(z^{2n}+\frac{1}{z^{2n}}+2(-1{)}^n\right)\right|$
$=\left|\sum _{n=1}^{15}{w}^{2n}+\frac{1}{w^{2n}}+2(-1{)}^n\right|$
$=\left|\frac{w^2\left(1-w^{30}\right)}{1-w^2}+\frac{\frac{1}{w^2}\left(1-\frac{1}{w^{30}}\right)}{1-\frac{1}{w^2}}+2(-1)\right|$
$=\left|\frac{w^2(1-1)}{1-w^2}+\frac{\frac{1}{w^2}(1-1)}{1-\frac{1}{w^2}}-2\right|$
$=|0+0-2|=2$