# Imaginary parts of roots of unity

Hi all,

What happens when we take the product of the imaginary parts of all the n-roots of unity (excluding 1)?
I read somewhere that we get n/(2^(n-1)).
How can we prove this?

Thanks!

Eh, your problem statement is a little faulty, I think. Anyways, here is a neat derivation I learned awhile ago:

By DeMoivre, the nth roots of unity are given by

$$\varepsilon _k = \cos{\frac{2k\pi}{n} + i\sin{\frac{2k\pi}{n}},$$

where k = 0, 1, ..., n-1. Geometrically, we can think of the roots of unity as the vertices of a regular n-gon inscribed in the unit circle.

Another way to think about this is to relate the roots of unity to the defining polynomial $f(z) = z^n - 1 = (z-1)(z-\varepsilon)\cdot\cdot\cdot(z-\varepsilon ^{n-1})$, where
$$\varepsilon = \cos{\frac{2\pi}{n} + i\sin{\frac{2\pi}{n}}.$$

It is clear how this is related to the nth roots of unity given above. If we take the derivative of both sides the equation relating f(z) to the product of factors, we get $nz^{n-1}$ on the one hand, and on the the hand, we have $$(z-\varepsilon)(z-\varepsilon ^2)\cdot\cdot\cdot(z-\varepsilon ^{n-1}) + \mbox{ sums each consisting of products which include the factor (z-1) }$$

by the product rule. Consequently,

$$n = f'(1) = (1-\varepsilon)(1-\varepsilon ^2)\cdot\cdot\cdot(1-\varepsilon ^{n-1}).$$(***)

Now

$$1-\varepsilon ^k = 1 - \cos{\frac{2k\pi}{n} - i\sin{\frac{2k\pi}{n}}.$$

By the pythagorean identity, $\cos 2x = \cos^2 x - \sin^2 x$, and $\sin 2x = 2 \sin x \cos x,$ this last expression is equal to

$$2sin^2{\,\frac{k\pi}{n}} - 2i\sin{\frac{k\pi}{n}}\cos{\frac{k\pi}{n}} = 2\sin{\frac{k\pi}{n}}\left(\sin{\frac{k\pi}{n}}-i\cos{\frac{k\pi}{n}}\right).$$

Taking the modulus, we find that $|1-\varepsilon ^k| = 2\sin{\frac{k\pi}{n}}$ for k = 1, ..., n-1. Relating this to the equation at (***), we find a product that does give us n/(2^(n-1)), but it's not quite the product of the imaginary parts of the roots of unity.

Let $1, z_1, \dots z_{n-1}$ be the roots of unity.

We want to find :
$$\prod_{i=1}^{n-1} \frac{1}{2i} (z_i - \frac{1}{z_i}),$$
or
$$\left(\frac{1}{2i}\right)^{n-1} \times (-1)^{n-1} \prod_{i=1}^{n-1} \frac{(1 -z_i)(1+z_i)}{z_i} .$$

Now, if $P(z)=z^{n} -1$, then, for $z \neq 1$
$$\frac{P(z)}{z-1} = \prod_{i=1}^{n-1} (z-z_i) = 1+z+z^{2} + \dots z^{n-1}.$$
Hence,
$$\prod_{i=1}^{n-1} (1-z_i) = \lim_{z\to 1} P(z)/(z-1) = n.$$
Similarly,
$$\frac{P(-1)}{-2} = \prod_{i=1}^{n-1} (-1-z_i) = (-1)^{n-1} \prod_{i=1}^{n-1} (1+z_i).$$.

Hence,
$$\prod_{i=1}^{n-1} (1+z_i) = \frac{(-1)^{n-1}+1}{2}.$$.

And,
$$\prod_{i=1}^{n-1} z_i = (-1)^{n-1}.$$.

So I get the answer as:

$$\left(\frac{1}{2i}\right)^{n-1} \times n \times \frac{(-1)^{n-1}+1}{2}.$$.