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This is a problem from G.Stephenson's Mathematical Methods for Science Students which I am stuck on:

Prove that, if $z = \cos \theta + i \sin \theta$ and $ n $ is any positive integer, $ z^n - \frac {1}{z^n} = 2i \sin n \theta$

Show that $$ z^{2n} -1= (z^2-1) \prod_{r=1}^{n-1} (z^2 -2z\cos \frac {r\pi}{n} + 1)$$

By substituting $z = \cos \theta + i \sin \theta$, deduce that

$$\frac {\sin n\theta}{\sin \theta} = 2^{n-1} \prod_{r=1}^{n-1} (\cos\theta - \cos \frac {r\pi}{n}) $$

Attempts at the solution:

The first part is standard, but beyond that I'm stuck. Presumably I have to show that $$2iz^n\sin n \theta = (z^2-1) \prod_{r=1}^{n-1} (z^2 -2z\cos \frac {r\pi}{n} + 1)$$ but I don't know how. I can't figure out the product term and the $\cos \frac {r\pi}{n}$ term in it. It looks something like the real part of an nth root of $\pm 1$ but I don't know what I'm supposed to do with that.

Any help (hints) would be appreciated

EDIT: I think I've seen something that will be important in solving the problem, but I'm still stuck for some cases of $ n $.

The basic idea is to group the $ r $ amd $ n-r $ terms, since$ \cos \frac {r\pi}{n} = -\cos \frac {n-r\pi}{n}$. Using $ (a-b)(a+b) = a^2 - b^2$, we can reduce the product term to

$$ \prod_{r=1}^{\frac {n-1}{2}} (z^4 -2z^2\cos \frac {2r\pi}{n} + 1) $$ for odd $ n $, and

$$ (z^2+1)\prod_{r=1}^{\frac {n-2}{2}} (z^4 -2z^2\cos \frac {2r\pi}{n} + 1) $$ for even $ n $

My intuition now is to continue reducing the product term in this manner until we are left with one or two terms which we can show, when multiplied with $ z^2-1$, equals the left hand side. However, we clearly meet a problem whenever $n$ has an odd factor, since you can't group the terms $ \cos \frac {2r\pi}{n} $ when $ n $ is odd. In effect, then, I have proven the identity for $ n=2^a $, $ a $ being some positive integer. So how do I proceed when $n $ is odd or has an odd factor?

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After reading the first few lines my first intuition would be to try considering the auxiliary function $f(z) = z^{n} - \frac{1}{z^{n}} - 2i\sin(n\theta)$ and show that this is a bounded entire function and hence a constant, then show that the constant must be zero. – JessicaK Apr 8 '14 at 7:35
I have no issue with proving $ z^n - \frac {1}{z^n} = 2i \sin (n\theta) $ which is simple enough using de Moivre's. I'm stuck at proving the stuff that comes after :S – user141554 Apr 8 '14 at 8:23

Call your LHS $f(z)$ and the RHS $g(z)$, both of which are polynomials of degree $2n$ in $z$. You are required to prove these two polynials agree on all complex numbers of modulus 1; this means the polynomial $f(z)-g(z)$ has all numbers of modulus 1 as its roots; this forces $f(z)\equiv g(z)$.

For this purpose it suffices to show both $f(z)$ and $g(z)$ have the same set of roots (they are already normalized, that is highest coefficient is 1). For $f(z)$ all $2n$-th roots of unity are the roots. In the rhs you have its factorization into quadratics with the roots grouped in pairs of reciprocals; $(e^{2\pi ir/2n},\ r=0,1,\dots 2n-1$ being the roots).

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Thank you for your answer! Of course, it seems so obvious now that you point out the quadratic can be factorized (so my point about the cos term being a root of unity was apparently the right way to go). This always seems to be the case for proving questions :S when you 'see' it it all falls in place. (Oh, and by reciprocals do you mean conjugates? ) – user141554 Apr 9 '14 at 14:52
For any complex number of modulus 1 its reciprocal and conjugate are the same. – P Vanchinathan Apr 9 '14 at 15:04

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