Sum of Complex Roots of Unity

Given:

$$x_k = \cos\left(\frac{2\pi k}{n}\right) + i \sin\left(\frac{2\pi k}{n}\right) = e^{2\pi i k/n}$$

Required: Find: $$\sum_{k=1}^{n} x_k$$

This is the sum of all $n^\text{th}$ roots of unity (from $k = 1$ to $n$).

We know: $$\sum_{k=0}^{n-1} e^{2\pi i k/n} = 0$$ So shifting index from $k = 1$ to $n$ just cycles the same roots: $$\sum_{k=1}^{n} e^{2\pi i k/n} = 0$$

✅ Final Answer:   $\boxed{0}$