To prove $\sum\limits_{k=1}^{n}k \cos{\frac{2k\pi}{n}} =-\frac {n}2$

Question

Prove that $\sum\limits_{k=1}^{n}k \cos{\frac{2k\pi}{n}} =-\frac {n}2$, for $n\in\mathbb{Z}, n\ge3$

Answer 1

Hint : $$\sum_{k=1}^nk\cos\left(\frac{2k\pi}{n}\right)=\Re\left[\sum_{k=1}^n ke^{\frac{2ik\pi}{n}}\right]=\Re\left[\sum_{k=1}^n k\left(e^{\frac{2i\pi}{n}}\right)^k\right]=\Re\left[e^{\frac{2i\pi}{n}}\sum_{k=1}^n k\left(e^{\frac{2i\pi}{n}}\right)^{k-1}\right]=\Re\left[e^{\frac{2i\pi}{n}}\frac{d}{dx}\left(\sum_{k=1}^n x^k\right)\Bigg|_{x=e^{\frac{2i\pi}{n}}}\right]=\cdots$$

Answer 2

$$1+x+x^2+\cdots+x^n=\frac{1-x^{n+1}}{1-x}$$ Differentiating both sides with respect to $x$, $$1+2x+3x^2+\cdots+nx^{n-1}=\frac{d}{dx}\frac{1-x^{n+1}}{1-x}$$ $$\therefore (1+x+\cdots+x^{n-1})+\sum_{k=1}^{n-1}kx^k=\frac{d}{dx}\frac{1-x^{n+1}}{1-x}$$ $$\therefore\sum_{k=1}^nk\left(e^{\frac{2i\pi}{n}}\right)^k=\left[\frac{d}{dx}\frac{1-x^{n+2}}{1-x}\right]_{x=e^{\frac{2i\pi}{n}}}-\sum_{k=1}^n\left(e^{\frac{2i\pi}{n}}\right)^k$$ The rest is very easy, just note that $e^{\frac{2i\pi}{n}}$ is the $n$th root of unity and $e^{i\theta}=\cos\theta+i\sin\theta$, and equate the real parts on both sides of the equality.