If $\displaystyle z_1,z_2,z_3$ be the complex numbers represented by the vertices $\displaystyle A, B, C$ of a triangle $\displaystyle \mbox{ABC}$ in anticlockwise order, show that:

$\displaystyle \left|\sum \frac{z_2 - z_3}{z_1}|z_1|^2\right|\cos A - 2R[(z_1 - z_2)(\bar{z_1} - \bar{z_2})]\sin A = 0$