Question

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

Answer 1

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