Complex roots of a 7th degree polynomial

Relevant wiki: Roots of Unity

The given polynomial is a finite geometric progression . It can be rewritten as:

$\frac{x^8-1}{x-1}=0$ Alternatively, $x^8-1=0\text{, }x\ne1$

Thus, the solutions of the equation will be all the 8 $^\text{th}$ roots of unity except for $1$ .

These solutions are found with $x_k=e^{2k\pi i/8}$ for $k=1,2,3,4,5,6,7$ .

On the complex plane, these solutions are located on the unit circle in $\frac{\pi}{4}$ increments:

The problem asks to find the sum of the squares of the real parts. There are four solutions in which the square of the real part is $\frac{1}{2}$ , and there is one solution in which the square of the real part is $1$ . The remaining two solutions do not have real parts.

Thus, the sum of the squares of the real parts is $\boxed{3}$ .

matlab code:

y=roots(ones(1,8));

x=real(y);

sum=0;

for i=1:size(x,1)

sum=sum+((x(i,1))^2);

end

sum