Solution: Proving $s = 0$ for the Sum of $n$-th Roots of Unity

Let $U_n = \{ e^{2\pi i k / n} \mid k = 0, 1, \dots, n-1 \}$, the set of all $n$-th roots of unity. We want to show that the sum of all elements $s = \sum_{k=0}^{n-1} e^{2\pi i k / n} = 0$.

Using the Geometric Series Formula

1. Let $c = e^{2\pi i / n}$, the primitive $n$-th root of unity. Then: $s = 1 + c + c^2 + \cdots + c^{n-1}.$ This is a finite geometric series.

2. Recall the sum of a geometric series: $1 + r + r^2 + \cdots + r^{m-1} = \frac{1 - r^m}{1 - r}, \quad r \neq 1.$

3. Here, $r = c$, and since $c^n = e^{2\pi i} = 1$ (the defining property of $n$-th roots of unity), we have: $s = \frac{1 - c^n}{1 - c}.$

4. Substituting $c^n = 1$: $s = \frac{1 - 1}{1 - c} = 0.$

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

---

Alternative Proof: Symmetry Argument

1. Multiplying $s$ by $c$, we rearrange the elements of the sum: $c \cdot s = c \cdot (1 + c + c^2 + \cdots + c^{n-1}) = c + c^2 + c^3 + \cdots + c^n.$

2. Because $c^n = 1$, this can be rewritten as: $c \cdot s = c + c^2 + \cdots + c^{n-1} + 1 = s.$

3. Subtract $s$ from both sides: $c \cdot s - s = 0 \implies (c - 1) \cdot s = 0.$

4. Since $c \neq 1$, it follows that: $s = 0.$

---

Conclusion

The sum of all $n$-th roots of unity is $s = 0$.

---

Would you like a deeper explanation, or clarification on any part?

---

Related Questions:

1. How are the $n$-th roots of unity distributed geometrically in the complex plane?
2. Why does the condition $c^n = 1$ guarantee a complete cycle of roots?
3. How can this property be used in solving polynomial equations?
4. Can you prove the sum $s = 0$ using symmetry arguments in the complex plane directly?
5. How does the geometric sum formula generalize for roots of unity?

Tip:

When working with roots of unity, visualizing them on the unit circle often clarifies their symmetry and properties.