What is the Sum of nth Roots of Unity and How Can It Be Proven?

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Homework Help Overview

The discussion revolves around proving that the sum of the nth roots of unity equals zero. The subject area includes complex numbers and polynomial equations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the definition of roots of unity and the manipulation of the sum S. There is an exploration of geometric series and attempts to manipulate S to demonstrate that it equals zero.

Discussion Status

Some participants are questioning the initial setup and notation regarding the roots of unity. There is an indication that corrections have been suggested, and some guidance has been offered regarding the manipulation of S. Multiple interpretations of the approach are being explored.

Contextual Notes

There are mentions of potential mistakes in the setup of the roots and the manipulation process. The discussion reflects uncertainty about the correct approach and the assumptions being made regarding the series and its properties.

converting1
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i'm trying to prove the sum of nth roots of unity = 0, but I don't really know how to proceed:

suppose z^n = 1 where z ε ℂ,

suppose the roots of unity for z are 1, ω, ω^2, ω^3 ... ω^n

the sum of these would be S = 1 + ω, ω^w, ω^3 +...+ ω^(n-1) + ω^n

from here I had an idea to do some fancy manipulation of S, then show that S = 0, but if say I do ωS - S I don't get 0!

I'm assuming I've made a very silly mistake or the way of approaches this is all wrong,

does anyone have a better approach or can anyone spot my mistake?

thanks,
 
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converting1 said:
i'm trying to prove the sum of nth roots of unity = 0, but I don't really know how to proceed:

suppose z^n = 1 where z ε ℂ,

suppose the roots of unity for z are 1, ω, ω^2, ω^3 ... ω^n
That's too many. Your roots should start at exponent 0 and end at exponent (n - 1).
converting1 said:
the sum of these would be S = 1 + ω, ω^w, ω^3 +...+ ω^(n-1) + ω^n
S = 1 + ω + ω2 + ... + ωn - 1

Notice that the right side is a (finite) geometric series.
converting1 said:
from here I had an idea to do some fancy manipulation of S, then show that S = 0, but if say I do ωS - S I don't get 0!

I'm assuming I've made a very silly mistake or the way of approaches this is all wrong,

does anyone have a better approach or can anyone spot my mistake?

thanks,
 
And if you make the correction Mark44 gave, you will find ωS - S is zero.
 
alternatively, they're the n roots of the polynomial equation xn - 1 = 0 …

so which coefficient is the sum of the roots? :wink:
 

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