If ${{x}}_k=\cos \Bigg{(}\frac{2\pi k}{n}\Bigg{)}+i\sin \Bigg{(}\frac{2\pi k}{n}\Bigg{)}$ , then $\sum ^n_{k=1}({{x}}_k)=?$

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:   \( \boxed{0} \)