If x_k=\cos \Bigg(2π k/n\Bigg)+i\sin \Bigg(2π k/n\Bigg) , then ∑ ^n_k=1(x_k)=?

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Question Id : 11534 | Context :NIMCET 2023

${{x}}_k=\cos \Bigg{(}\frac{2\pi k}{n}\Bigg{)}+i\sin \Bigg{(}\frac{2\pi k}{n}\Bigg{)}$ , then

$\sum ^n_{k=1}({{x}}_k)=?$

🎥 Video solution / Text Solution of this question is given below:

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: $\boxed{0}$

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