# Why isnt product of nth roots of unity always 1

I know product of nth roots of unity is 1 or -1 depending whether n is odd or even. But in this way I am getting 1. Where am I wrong?

$\text{Let }\alpha = \cos \frac{2 \pi}{n} + \iota \sin \frac{2 \pi}{n} \text{ be a root of }x^n=1 \\ \text{Then product of nth roots will be } 1\cdot \alpha \cdot \alpha^2 ... \alpha^{n-1} = \alpha^{\frac{(n)(n-1)}{2}}\\ =\left( \alpha^n \right)^{\frac{n-1}{2}}\\ =1^{\frac{n-1}{2}} \text{......By the definition of alpha ??}\\ =1$

I can see this doesn't even work for n=2.

• Using a similar logics, $$-1=(-1)^{2/2}=((-1)^2)^{1/2}=1^{1/2}=1.$$
– user65203
Commented May 16, 2017 at 12:29
• My detailed answer here Commented Apr 22, 2021 at 17:55

Consider the following: $$(-1)^1 = (-1)^{2\cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}=1^{\frac{1}{2}} = 1$$ You did something like that. $n(n-1)/2$ is always an integer when $n$ is a natural number, but when you take out the $n$ you are left with $(n-1)/2$ which is not always an integer, which messes things up when $n$ is even. In short, the rule $a^{bc} = (a^b)^c$ does not always hold for complex numbers, in particular when $b$ or $c$ are not integers.

• As you showed by your example, $a^{bc} = (a^b)^c$ doesn't even always hold for real numbers
– user304329
Commented May 16, 2017 at 12:28

$(n-1)/2$ is not an integer, so $1^{(n-1)/2}$ involves the square-root, which might be positive or negative. You have to look more closely to decide which.
It is a bit like $-1=(-1)^{2/2}=1^{1/2}=1$

Hint:

$$\alpha^{n(n-1)/2}=(\alpha^{n/2})^{n-1}$$

Now $\alpha^{n/2}=\cdots=-1$