The centre of a square ABCD is at orign and the point A is represented by $z_1$  The centroid of triangle BCD is represented by

Given A is represented by $z_1.$  Since the diagonals of a square bisect cach other at right angles, therefore,$\angle AOB=\angle BOC=\angle COD=\frac{\pi }{2}$

And hence $B\equiv i{z}_1,c\equiv i^2{z}_1$  and $D\equiv i^3{z}_1$

$\therefore$  Centroid of triangle BCD is represented by

$\frac{i{z}_1-z_1-i{z}_1}{3}=-\frac{z_1}{3}.$