x2013-1=0=(x-1)(x2012+x2011+x2010+...+x2+x+1). Since x≠1:
x2012+x2011+x2010+...+x2+x+1=0=∑xk for 0≤k≤2012 (1)
xn=einθ=cos(nθ)+isin(nθ), so if x2013=1, cos(2013θ)+isin(2013θ)=1.
This means that sin(2013θ)=0 and cos(2013θ)=1⇒θ=0 and x=1, or θ=2π/2013.
On an Argand diagram the 2013 solutions would be represented by the endpoints of the radii of a regular polygon with 2013 sides. Each radius has unit length and the first root would be the point (1,0) which has no imaginary component, that is, x=1. The next root would be (cos(2π/2013), sin(2π/2013)), the next (cos(4π/2013), sin(4π/2013)), then (cos(6π/2013), sin(6π/2013)), and so on up to (cos(4024π/2013), sin(4024π/2013)), and then (cos(4026π/2013), sin(4026π/2013))=(cos(2π),sin(2π))=(1,0) which is where we started. So there are 2012 complex roots. 2π/2013 is about 0.00312 radians or about 0.179°, a tiny angle. The polygon would almost resemble a perfect unit circle.
xk=e2πki/2013 and ∑xk=∑e2πki/2013=0 for 0≤k≤2012 by substitution into (1).