Step-by-step Solution:

The expression you provided relates to the sum of the \(n\)-th roots of unity in complex numbers. \[ x_k = \cos \Bigg{(}\frac{2\pi k}{n}\Bigg{)}+i\sin \Bigg{(}\frac{2\pi k}{n}\Bigg{)} = e^{i \frac{2 k \pi}{n}} \] This is the general form of the \(n\)-th roots of unity. \[\] Summing up the \(n\)-th roots of unity: \[ \sum_{k=1}^{n} x_k = \sum_{k=1}^{n} e^{i \frac{2 k \pi}{n}} \] This is a geometric series: \[ e^{i \frac{2 \pi}{n}} + e^{i \frac{4 \pi}{n}} + e^{i \frac{6 \pi}{n}} + \ldots + e^{i \frac{2 n \pi}{n}} \] Let: \[ \alpha = e^{i \frac{2 \pi}{n}} \] Thus, the sum becomes: \[ \alpha + \alpha^2 + \alpha^3 + \ldots + \alpha^n \] This is the sum of a geometric series with the first term \(\alpha\) and common ratio \(\alpha\), and there are \(n\) terms. The sum of a geometric series is given by: \[ \text{Sum} = \frac{\alpha(1 - \alpha^n)}{1 - \alpha} \] Substituting \(\alpha = e^{i \frac{2 \pi}{n}}\), we get: \[ \frac{e^{i \frac{2 \pi}{n}}(1 - e^{i 2 \pi})}{1 - e^{i \frac{2 \pi}{n}}} \] Now, since \(e^{i 2 \pi} = 1\), we have: \[ 1 - e^{i 2 \pi} = 1 - 1 = 0 \] So the entire expression simplifies to: \[ \frac{e^{i \frac{2 \pi}{n}} \cdot 0}{1 - e^{i \frac{2 \pi}{n}}} = 0 \] The sum of all the \(n\)-th roots of unity is zero: \[ \sum_{k=1}^{n} e^{i \frac{2 k \pi}{n}} = 0 \]