Shouldn't the product of all $n$-th roots of unity be $1$?

Question

nth root of unity:

$$ 1^{1/n} = e^{i2\pi k/n}\ \ \ \ k=0,1,...,n-1 $$

multiplying together all the nth roots of unit to get back "1":

$$ 1=\prod_{k=0}^{n-1}e^{i\left(2\pi k/n\right)} $$

for n=2:

$$ 1=\left(e^{i2\pi\left(0\right)/2}\right)\left(e^{i2\pi\left(1\right)/2}\right) \\ 1=\left(1\right)\left(-1\right) \\ 1=\ -1 $$

why is 1 = -1?

For n=4:

$$ 1=\left(e^{i2\pi\left(0\right)/4}\right)\left(e^{i2\pi\left(1\right)/4}\right)\left(e^{i2\pi\left(2\right)/4}\right)\left(e^{i2\pi\left(3\right)/4}\right) \\ 1=\left(1\right)\left(i\right)\left(-1\right)\left(-i\right) \\ 1=\ -1 $$

why is 1 = -1? shouldn't it be "1"?

A few definitions:

$$ e^{i\theta} = cos(\theta) + i sin(\theta) $$

$$ e^{-i\theta} = cos(\theta) - i sin(\theta) $$

$$ i*i = -1 $$

$$ \frac{1}{i} = -i $$

Answer 1

The $n$th roots of unity are the roots of the polynomial $x^n - 1$.

If you factor this polynomial as $$x^n - 1 = \prod_{j=0}^{n-1} (x-e^{2\pi i j/n}) ,$$ then you will notice by multiplying out and comparing constant terms that the product of the roots is $\pm 1$, with the sign depending on whether $n$ is even or odd.

Answer 2

Clearly the original claim is false.

Roots of unity come in two types: the real ones, and the pairs of complex conjugates. The complex conjugates, being of unit modulus, give pairwise products of $1$. That leaves the real roots. For an odd index like cube roots there is just the real root $1$, which when multiplied by all those complex conjugate pairs still gives $1$. But for even indices like square or fourth roots, there's that other real root $-1$. That's when the overall product becomes $-1$, too.