Mathematics - Complex Number Question with Solution | TestHub

Given, $z^2+z+1=0\Rightarrow z=w,w^2$

Now, $\left|\sum _{n=1}^{15}{\left(z^{\mathrm{n}}+\left(-1\right)\frac{1}{z^n}\right)}^2\right|=\left|\sum _{n=1}^{15}\left(z^{2n}+\frac{1}{z^{2n}}+2{\left(-1\right)}^n\right)\right|$

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

$=\left|\frac{w^2\left(1-w^{30}\right)}{1-w^2}+\frac{\frac{1}{w^2}\left(1-\frac{1}{w^{30}}\right)}{1-\frac{1}{w^2}}+2\left(-1\right)\right|$

$=\left|\frac{w^2\left(1-1\right)}{1-w^2}+\frac{\frac{1}{w^2}\left(1-1\right)}{1-\frac{1}{w^2}}-2\right|$

$=\left|0+0-2\right|=2$