Prove that the nth root of unity form a geometric progression .also show that the sum of those n roots is zero and their product is(-1)^n-1

                             BS GREWAL EXERCISE QUESTION 

Question :-Prove that the nth root of unity form a geometric progression .also show that the sum of those n roots is zero and their product is(-1)^n-1

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Solution:- let x is the nth root of unity form a series of geometric progression ,we have

x^n=1

x=(1)^1/n

x=(1+i0)1/n

x=(cos0+isin0)^1/n=(cos2rπ+isin2rπ)^1/n

x=cis2rπ/n

Putting r=0,1,2,3,3…,n-1

x=1,(cis2π/n)^r=w^r

.i.e.  x=1,w,w^2,w^3,……..,w^(n-1)

Clearly its form a geometric progression since w is one nth root of unity

Then w^n=1

w^n-1=0

(w-1)(1+w+w^2+…….+w^n-1)=0

w≠ 1

(1+w+w^2+…….+w^n-1)=0

Hence the sum of nth root of unity is zero

Again, 1*w*w^2.……..*w^n-1

w^(1+2+3+4…n-1)

w^(n(n-1)/2)

[(Cisπ)^2/n]^n(n-1)/2

(-1)^n-1

 

 hence proved😉😋

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