Question

Can someone check if this is right:
Need to prove that the sum of distinct nth roots of unity is equal 0.
Looks to simple to me....
Proof: The different n rooths of unity can be seen like vertices of an n-gon inscribed in a unit circle of a complex plane. This n-gon is invariant under the rotation by 2Pi/n. In the complex numbers this rotation is expressed by multiplication by
e^i(2Pi/n)
So if the sum befor rotation is S, after rotation should be same. It means S = Se^i(2Pi/n) giving the only posibility that S =0

Answer 1

hi @experimentX

Answer 2

It convinces me.

Answer 3

ok, thx for help @phi

Answer 4

does anybody know how to prove this algebraicly?

Answer 5

hello myko

Answer 6

the yesterdays problem was hyperbola :)

Answer 7

yeah .. i saw sam's solution ..!