# Thread: Complex numbers and trigonometric result?

1. ## Complex numbers and trigonometric result?

By using the concepts of complex numbers, evaluate $\sum_{k = 1}^{3} \cos \frac{2k\pi}{7}$

2. Let $z=\cos\frac{\pi}{7}+i\sin\frac{\pi}{7}\Rightarrow z^7=-1\Rightarrow(z+1)(z^6-z^5+z^4-z^3+z^2-z+1)=0$

We have: $\cos\frac{2\pi}{7}=\frac{1}{2}\left(z^2+\frac{1}{z ^2}\right)=\frac{z^4+1}{2z^2}$

$\cos\frac{4\pi}{7}=\frac{z^8+1}{2z^4}=\frac{1-z}{2z^4}$

$\cos\frac{6\pi}{7}=\frac{z^{12}+1}{2z^6}=\frac{1-z^5}{2z^6}$

Then $\frac{z^4+1}{2z^2}+\frac{1-z}{2z^4}+\frac{1-z^5}{2z^6}=\frac{-z^5+z^4-z^3+z^2-z+1}{2z^6}=\frac{-z^6}{2z^6}=-\frac{1}{2}$