Question

If $z_1,z_2,z_3,z_4,z_5$ are roots of the equation $z^5+z^4+z^3+z^2+z+1=0$, then the value of $\left|\sum _{i=1}^5{z}_i^4\right|$ is

Answer 1

$z^5+z^4+z^3+z^2+z+1=0\Rightarrow (z^5+z^4+z^3+z^2+z+1)(z-1)=0\Rightarrow z^6-1=0\Rightarrow z^6=1=e^{i2n\pi }\Rightarrow z=e^{i2n\pi /6},n=0,1,2,3,4,5$

$z_0=1,z_1=e^{i\pi /3},z_2=e^{i2\pi /3},z_3=e^{i\pi },z_4=e^{i4\pi /3},z_5=e^{i5\pi /3}$

Therefore,
$\sum _{i=0}^5{z}_i^4=1+e^{i4\pi /3}+e^{i8\pi /3}+e^{i4\pi }+e^{i16\pi /3}+e^{i20\pi /3}=1+(-\cos \frac{\pi }{3}-i\sin \frac{\pi }{3})+(-\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})+1+(-\cos \frac{\pi }{3}-i\sin \frac{\pi }{3})+(-\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})\Rightarrow \sum _{i=0}^5{z}_i^4=0\Rightarrow \sum _{i=1}^5{z}_i^4=-1$