Question

Let {z0, z1, z2, . . . , z2017} be the solution set of the equation z^2018 = 1. Show that z0 +z1 +z2 +···+z2017 = 0.

Answer 1

Consider the number z0 = exp(2*pi*i/5). This is one of the five
fifth-roots of unity. These five roots lie on the unit circle in the 
complex plane, equally spaced, and have the form

  z0, z0^2, z0^3, z0^4, 1

t is the similarity of these roots to the terms in the z-polynomial
that gives us a clue as to how to proceed. We also know from geometry 
that the vector sum of the five fifth-roots is zero (this is the key 
realization that saves us a huge amount of algebraic work). Let's just 
plug the number z0 into your original z-polynomial and use the fact 
that the sum of the five fifth roots is zero.  We see that

  1 + z0 + z0^2 + z0^3 + z0^4 = 0    

is equal to zero, so that (z-z0) = (z - exp(2*pi*i/5)) is one of the 
factors of the original equation. We could divide this factor into the 
original quartic z-polynomial and work with the cubic that remains, 
but we can continue with the same reasoning (and therefore have to 
perform less algebra). We consider the next fifth-root of unity:  
z1 = z0^2. If we plug this into the z-polynomial, we get

 1 + z1 + z1^2 + z1^3 + z1^4 = 1 + z0^2 + z0^4 + z0^6 + z0^8 
     = 1 + z0^2 + z0^4 + z0^1 + z0^3

because z0^5 = 1. Again, we see that this must be zero, and so we've
found another factor of the original z-polynomial:  

 (z-z1)=(z-z0^2) = (z-exp(4*pi*i/5)). 

Similarly, you can verify that z0^3 and z0^4 are also roots of the
original polynomial, and we obtain the final factorization as

 z^4 + z^3 + z^2 + z + 1 = (z-z0)(z-z0^2)(z-z0^3)(z-z0^4)

where z0 = exp(2*pi*i/5). You can take this number and plug it back in 
and multiply everything out to check that this equation is true.

I hope that helps! z0 +z1 +z2 +···+z2017 = 0.