Let $U_n$ denote the set of all the n-th roots of unity, i.e., the set of complex numbers of the form $c^k$ for $k = 0, 1, 2, \dots, n-1$, where $c = e^{2\pi i / n}$ is a primitive $n$-th root of unity. We want to prove that the sum of all elements in $U_n$, denoted by $s$, is zero, i.e.,

$s = 1 + c + c^2 + \cdots + c^{n-1} = 0.$

Step 1: Understanding the sum

The sum we need to consider is:

$s = 1 + c + c^2 + \cdots + c^{n-1}.$

This is a geometric series with the first term $1$ and the common ratio $c$, where $c = e^{2\pi i / n}$.

Step 2: Multiply the sum by $1 - c$

Now, let's multiply the entire sum by $1 - c$:

$(1 - c)s = (1 - c)(1 + c + c^2 + \cdots + c^{n-1}).$

Using the distributive property, we get:

$(1 - c)s = 1 + c + c^2 + \cdots + c^{n-1} - c - c^2 - c^3 - \cdots - c^n.$

Notice that all the terms in the sum cancel out except for the first and the last terms:

$(1 - c)s = 1 - c^n.$

Step 3: Simplify the expression

Since $c = e^{2\pi i / n}$, it follows that $c^n = e^{2\pi i} = 1$. Therefore,

$(1 - c)s = 1 - 1 = 0.$

This implies:

$(1 - c)s = 0.$

Step 4: Conclude that $s = 0$

Since $1 - c \neq 0$ (because $c$ is a nonzero complex number), we can conclude that:

$s = 0.$

Thus, the sum of all the $n$-th roots of unity is zero:

$1 + c + c^2 + \cdots + c^{n-1} = 0.$

Conclusion

The sum of all the $n$-th roots of unity, $s$, is indeed zero. This is a fundamental result in complex analysis and group theory, as the roots of unity are symmetrically distributed around the unit circle in the complex plane.

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Would you like more details on any step or further clarifications? Here are some related questions that might deepen your understanding:

1. Why are the n-th roots of unity symmetrically arranged on the unit circle in the complex plane?
2. How does the formula for the sum of a geometric series apply to this problem?
3. What is the significance of the n-th roots of unity in group theory?
4. Can this result be generalized to show the sum of roots of any polynomial equation?
5. How can the properties of the primitive roots of unity be used in Fourier analysis?

Tip: When dealing with sums of roots of unity, it's often useful to multiply the sum by something that simplifies the expression, such as $1 - c$ in this case.