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$$w^n=1$$ $$\omega^n-1=0$$ But $$\omega^n-1=(1-\omega)(1-\omega^2)....(1-\omega^{n-1}$$because $\omega, \omega^2....$ are roots of the the equation $\omega^n-1$

So the answer should be zero. But the answer given is ‘n’. What’s going wrong?

Aditya
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    Your equation is wrong. You are forgetting one of the $n$ roots on RHS. The given answer is also wrong. For $n=2$ then product is $1-w$. – Kavi Rama Murthy Feb 02 '20 at 07:25
  • For $n=2$, $\omega=-1$, so the answer is still $n=2=1-(-1)$. – Pythagoras Feb 02 '20 at 07:46
  • See https://math.stackexchange.com/questions/1909362/product-of-one-minus-the-tenth-roots-of-unity/1909366#1909366 and https://math.stackexchange.com/questions/806636/question-on-primitive-roots-of-unity – lab bhattacharjee Feb 02 '20 at 11:42

3 Answers3

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The polynomial $x^n-1$ has $n$ complex roots, given by the $n^{th}$ roots of unity, which are $$ 1,\omega,\omega^2,\ldots,\omega^{n-1}, $$ where $\omega=e^{2\pi i/n}$. Thus it has the factorization $$ x^n-1=(x-1)(x-\omega)(x-\omega^2)\ldots(x-\omega^{n-1}). $$ If you substitute $x=1$ on both sides, you obtain $0=0$, since $x=1$ is a root. If you want to obtain a non-trivial product when $x=1$, you first need to get rid of the $x-1$ on both sides, like so: $$ \frac{x^n-1}{x-1}=(x-\omega)(x-\omega^2)\ldots(x-\omega^{n-1}). $$ If you take the limit as $x\to 1$ in this equation you will find (using this rule) that $$ n=(1-\omega)(1-\omega^2)\ldots(1-\omega^{n-1}). $$

pre-kidney
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$x^n-1=(x-1)\cdots(x-w^{n-1})\implies x^{n-1}+x^{n-2}+\cdots+x+1= \dfrac{x^n-1}{x-1}=(x-w)\cdots(x-w^{n-1})$. Now plugging in $x=1$, we get $n$.

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y^n-1={1+y+y^2+…+y^(n--1)}(y-1)

then, (y-1)(y-ω)(y-ω^2)…{y-ω^(n-1)}=y^n-1={1+y+y^2+…+y^(n-1)}(y-1) Now take y=x+1 Therefore x{x+(1-ω)}{x+(1-ω^2)}…[x+{1-ω^(n-1)}] ={1+(x+1)+(x+1)^2…+(x+1)^(n-1)}x

Comparing coefficients of x in both sides of the above identity

(1-ω)(1-ω^2)(1-ω^3)…{1-ω^(n-1)}=1+1+1+…upto n=n

Is it a proof?

sitangsu
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