What is the product of all $n$th roots of unity?
Understand the Problem
The question is asking for the product of all the $n$th roots of unity. The $n$th roots of unity are the complex numbers that, when raised to the power of $n$, equal 1. We need to determine this product.
Answer
$(-1)^{n-1}$
Answer for screen readers
$(-1)^{n-1}$
Steps to Solve
- Represent the $n$th roots of unity
The $n$th roots of unity are the solutions to the equation $z^n = 1$. These roots can be expressed in the form $z_k = e^{2\pi i k/n}$, where $k = 0, 1, 2, ..., n-1$.
- Write the product of the roots
Let $P$ be the product of all $n$th roots of unity. Then, $$P = \prod_{k=0}^{n-1} e^{2\pi i k/n} = e^{\sum_{k=0}^{n-1} (2\pi i k/n)}$$
- Simplify the exponent
The sum in the exponent is an arithmetic series. We can simplify the exponent as follows: $$ \sum_{k=0}^{n-1} \frac{2\pi i k}{n} = \frac{2\pi i}{n} \sum_{k=0}^{n-1} k = \frac{2\pi i}{n} \cdot \frac{(n-1)n}{2} = \pi i (n-1) $$
- Substitute the exponent back into the expression for P
Substituting this back into the expression for $P$, we get: $$ P = e^{\pi i (n-1)} $$
- Simplify the exponential term
We know that $e^{i\pi} = -1$. Therefore, $$ P = e^{\pi i (n-1)} = (e^{i\pi})^{(n-1)} = (-1)^{n-1} $$
$(-1)^{n-1}$
More Information
The product of the $n$th roots of unity is either 1 or -1, depending on whether $n$ is odd or even. If $n$ is odd, then $n-1$ is even, so the product is 1. If $n$ is even, then $n-1$ is odd, so the product is -1.
Tips
A common mistake is incorrectly calculating the sum of the exponents or simplifying the final exponential expression. Another mistake could be not recognizing the $n$th roots of unity in exponential form.
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