The problem is to prove that the sum of all $n$-th roots of unity is zero, $s = \sum_{z \in U_n} z = 0$, where $U_n = \{ e^{2\pi i k / n} \mid k = 0, 1, \dots, n-1 \}$. Let’s use the hint and tackle this step-by-step.

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Solution:

Step 1: Represent the $n$-th roots of unity

Let $c = e^{2\pi i / n}$, which is a primitive $n$-th root of unity. Then the $n$-th roots of unity are: $U_n = \{ 1, c, c^2, \dots, c^{n-1} \}.$ We need to compute the sum: $s = 1 + c + c^2 + \dots + c^{n-1}.$

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Step 2: Observe the geometric series

The sum $s$ can be expressed as a geometric series: $s = 1 + c + c^2 + \dots + c^{n-1}.$ The sum of a geometric series with ratio $c \neq 1$ is given by: $s = \frac{1 - c^n}{1 - c}.$

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Step 3: Simplify using the property of $c^n$

Since $c = e^{2\pi i / n}$, it satisfies $c^n = e^{2\pi i} = 1$. Substituting $c^n = 1$ into the formula: $s = \frac{1 - 1}{1 - c} = \frac{0}{1 - c} = 0.$

Thus, the sum of all $n$-th roots of unity is $s = 0$.

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Step 4: Alternative explanation using symmetry

Another way to see this is that the $n$-th roots of unity are symmetrically distributed around the origin in the complex plane. The sum of vectors symmetrically distributed around the origin is always zero, as the contributions in all directions cancel out.

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Conclusion

The sum of all $n$-th roots of unity is: $s = \sum_{z \in U_n} z = 0.$

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Would you like me to elaborate on any part of the derivation or discuss more about the geometric properties of roots of unity?

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Related Questions:

1. Why does $c^n = 1$ for any primitive $n$-th root of unity?
2. How do the $n$-th roots of unity distribute on the unit circle?
3. Can this sum be generalized for a subset of $n$-th roots of unity?
4. What is the significance of the symmetry of roots of unity in other areas of mathematics?
5. How does the formula for the geometric series apply to complex numbers?

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Tip:

When working with roots of unity, remember their fundamental property: $c^k \cdot c^m = c^{k+m} \mod n$, and that they always lie on the unit circle in the complex plane!