Fdtxjr

$z_1,z_2,…z_n$ satisfy the equation $(z+1)^n=1$. Use the product of $z_1,z_2,…z_n$ to prove that
$$sin frac pin sin frac 2pin…sin frac (n-1)pin=frac n2^n-1$$

Attempt I compute $z_1=1-1=0,…z_n=cos frac(n-1)2pi n+isin frac(2n-1)pi n-1$

But if I do the multiplication ,the calculation result seems very ugly...

Hint. Note that
$$prod_k=1^n-1sinleft(frackpinright)=prod_k=1^n-1frace^ifrackpin-e^-ifrackpin2i=
frace^-ifracn(n-1)pi2n(2i)^n-1prod_k=1^n-1left(e^ifrac2kpin-1right)
=frac(-1)^n-1prod_k=2^nz_k2^n-1.$$
where
$$frac(z+1)^n-1z=z^n-1+nz^n-2+dots +n=(z-z_2)dots(z-z_n).$$
Can you take it from here?

Consider using the Viete's theorem.

We need to factor out your first root $z=0$. After that the constant term is $n$, and the highest term is $z^n-1$. So the product of the roots should be $(-1)^n-1n$.

But I can't relate the product to the main problem yet...

A different approach. Consider $x^2n-2x^ncos(na)+1=0 , ngt0 , , agt 0$. This gives you $x^n=exp pm i na$. or $x=exp pm ileft(a+frac2pi knright), k=0, 1, 2,ldots , n-1$.

$$beginalignedx^2n-2x^ncos(na)+1&=0\ prod_k=0^n-1left(x-exp ileft(dfraca+2pi knright)right)left(x-exp ileft(-dfraca+2pi knright)right)&=0\ prod_k=0^n-1left[ x^2-2xcosleft(a+dfrac2pi knright)+1right]&=0endaligned$$

Let $x=1$ and set $a=2b$. $$beginalignedprod_k=0^n-1left[1-2cosleft(2b+dfrac2pi knright)+1right]&=0\ prod_k=0^n-1left[2-2cos2left(b+dfracpi knright)right]&=0endaligned$$

$$beginalignedprod_k=0^n-1left[2-2cos2left(b+dfrackpinright)right]&=2-2cosleft(2nbright)equiv4sin^2left(nbright)endalignedtag1$$

Also $$beginaligned2-2cos2left(b+dfracpi knright)&=2-2left[cos^2left(b+dfracpi knright)-sin^2left(b+dfracpi knright)right]\&=4sin^2left(b+dfrackpi nright)endalignedtag2$$

Equate $(1)$ and $prod_k=0^n-1(2)$ and take square root of both sides to get the desired result.

$$beginalignedsqrt4sin^2left(nbright)&=sqrtprod_k=0^n-14sin^2left(b+dfrackpi nright)\2sin left(nbright)&=2^nprod_k=0^n-1sinleft(b+dfrackpi nright)\ dfracsinleft(nbright)sin b&=2^n-1dfracsinleft(bright)sin bsinleft(b+dfracpinright)cdotssinleft(b+dfrac(n-1)pinright)endaligned$$

Now take the limit as $bto 0^+$ which gives you: $$n=2^n-1underbracesinleft(dfracpinright)sinleft(dfrac2pinright)cdotssinleft(dfrac(n-1)pinright)_textLet it =varphiimplies boxedvarphi =dfracn2^n-1$$

Consider the expression $$(z+1) = Z $$
Then, $$Z^n=1 $$ Also, $$1=(-1)^2=e^i2kpi $$ Therefore, $$Z=(1)^1/n=e^i2kpi/n $$ And then, $$ z=e^i2kpi/n-1$$
Now, this has got $n$ roots for $ k =0 space to space n-1$ and we designate these roots as $z_0,z_1,z_2...z_n-1$. Now, coming back to original expression and expanding it:
$$beginalign*
(z+1)^n&=1\ z^n+nz^n-1+frac n2(n-1)z^n-2+..+nz+1&=1\ z^n+nz^n-1+frac n2(n-1)z^n-2+..+nz&=0
endalign*$$

$z=0$ is a root. So divide above equation by z and you will get:
$$beginalign*
z^n-1+nz^n-2+frac n2(n-1)z^n-3+..+n&=0
endalign*$$
The above equation now has roots $z_1,z_2,...,z_n-1$ and the product of these roots will be $(-1)^n-1n$ (ratio of coefficients of lowest power term to highest power term, positive for even degree polynomial and negative for odd degree polynomial).
$$prod_k=1^n-1left(e^ifrac2kpin-1right)=(-1)^n-1n $$

Now using the hint by Robert Z:

$$prod_k=1^n-1sinleft(frackpinright)=frac1(2i)^n-1prod_k=1^n-1left(e^ifrackpin-e^-ifrackpinright)=
frac1(2i)^n-1prod_k=1^n-1e^-ifrackpinleft(e^ifrac2kpin-1right).$$

And, $$frac1(2i)^n-1prod_k=1^n-1e^-ifrackpinleft(e^ifrac2kpin-1right) = frac(-1)^n-1n(2i)^n-1prod_k=1^n-1e^-ifrackpin = frac(-1)^n-1n(2i)^n-1e^-ifrac(n-1)pi2$$

Which on simplification, gives $$frac(-1)^n-1n(-2)^n-1 = fracn(2)^n-1$$