Prove that the area of a triangle whose vertices are the points $z_{1}, z_{2}$ and $z_{3}$ in the Argand's plane is $$\sum \frac{(z_2-z_3)|z_1|^2}{4iz_1}$$
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Area of a triangle with vertices $(x_k,y_k)$ $k=1,2,3$ is given by $$ A = \frac{1}{2}\begin{vmatrix}1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \end{vmatrix} $$ Now, if $z_k = x_k + iy_k$, then $$x_k = \frac{z_k + \bar{z_k}}{2}$$ and $$y_k = \frac{z_k - \bar{z_k}}{2i}$$ Substituting in the above determinant we obtain \begin{align*} A &= \frac{1}{8i}\begin{vmatrix}1 & z_1+\bar{z_1} & z_1-\bar{z_1} \\ 1 & z_2+\bar{z_2} & z_2-\bar{z_2} \\ 1 & z_3+\bar{z_3} & z_3-\bar{z_3} \end{vmatrix} \\ &=-\frac{1}{4i}\begin{vmatrix}1 & z_1 & \bar{z_1} \\ 1 & z_2 & \bar{z_2} \\ 1 & z_3 & \bar{z_3} \end{vmatrix} \end{align*} Now we get the result by expanding through the third column.