1. Complex numbers and triangles?

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

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

2. Clarity?

Just to clarify...

$R[a+bi]=a$

$\sum \frac{z_2 - z_3}{z_1}|z_1|^2$= $\frac{z_2 - z_3}{z_1}|z_1|^2+\frac{z_3 - z_1}{z_2}|z_2|^2+\frac{z_1 - z_2}{z_3}|z_3|^2$