Formula for area of triangle in complex plane

Question

If $A(z_1)$, $B(z_2)$, $C(z_3)$ are vertices of a triangle $ABC$ in Argand plane, what is the area of the triangle?

Answer 1

First assume $z_3 = 0$.

We need a degree 2 polynomial $A(z_1, z_2, \overline{z_1}, \overline{z_2})$ that is:

• bilinear (over $\mathbb{R}$) in $z_i$. Therefore of the form $P z_1 \overline{z_2} + Q \overline{z_1}z_2$.

• rotation invariant. This is true for any sum of $z \overline{z'}$ terms, so that requirement has been satisfied.

• negated by complex conjugation of both $z_i$ together. Therefore $P + Q=0$.

• negated by exchange of indices $1 \leftrightarrow 2$. This also means $P+Q=0$.

The area formula is then (constant) x ($ z_1 \overline{z_2} - \overline{z_1}z_2$).

For the triangle with vertices $1,i,0$ the constant is $-\frac{1}{4i}$.

The general formula is the same thing applied to $(z_1 - z_3, z_2 - z_3)$.

A more symmetric writing of the result is to take a cyclic ordering of 1,2,3 and then $A = (\sum \pm z_m \overline{z_n})/4i$ summed over all 6 ordered pairs of distinct $m,n$ with the $+$ sign for pairs in order, and $-$ for pairs out of order.