To make it short I have a doubt in the last step of a complex number inequality, the problem is the next one
Let complex numbers $z_1,z_2,z_3,...,z_n$ all modulus $1$ and $z_1+z_2+z_3+...+z_{2012}=0$ Show that. $$ \sum^{2012}_{k=1}\vert z-z_k \rvert \geq 2012 \ \forall z \in \mathbb{C}$$
The last step given in the book states that $\vert \sum_{2012}^{k=1}(z\overline{z_k})-2012 \vert=2012$ However I cannot prove this, I suppose that if we somehow show that $\overline{z_1}+\overline{z_2}+\overline{z_3}+...+\overline{z_{2012}}=0$ then we would be done, However the only thing that ocurred to me is to manipulate the given expression $z_1+z_2+z_3+...+z_{2012}=0 \Rightarrow \frac{1}{\overline{z_1}}+\frac{1}{\overline{z_2}}+\frac{1}{\overline{z_3}}+....+\frac{1}{\overline{z}_{{2012}}}=0 $ But then we would have a nasty expression. Also the fact that the complex numbers sum to zero reminds of roots of unity but I didn't see a conection.