What is the area of the triangle having $z_1$, $z_2$ and $z_3$ as vertices in Argand plane?

Question

What is the area of the triangle having $z_1$, $z_2$ and $z_3$ as vertices in Argand plane? Is it $$\frac{-1}{4i}[z_1(z_2^* - z_3^*)-z_1^*(z_2-z_3)+{z_2(z_3^*)-z_3(z_2^*)}]$$ where $w^*$ denotes the complex conjugate?

Answer 1

Let $z_j = x_j + iy_j$, $j = 1, 2, 3$. The area of the triangle is given by \begin{align*} \frac{1}{2} \begin{vmatrix} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ 1 & x_3 & y_3 \end{vmatrix}&= \frac{1}{2} \begin{vmatrix} 1 & x_1+iy_1 & y_1 \\ 1 & x_2+iy_2 & y_2 \\ 1 & x_3+iy_3 & y_3 \end{vmatrix}\\ &= \frac{1}{4i} \begin{vmatrix} 1 & z_1 & z_1-z_1^*\\ 1 & z_2 & z_2-z_2^* \\ 1 & z_3 & z_3-z_3^* \end{vmatrix}\\ \end{align*} Now expand via first column to get the required expression.

Answer 2

If ${\bf 0}=(0,0)$, ${\bf z}_1=(x_1,y_1)$, and ${\bf z}_2=(x_2,y_2)$ are the vertices of a triangle $\triangle$ in the $(x,y)$-plane then the signed area of $\triangle$ is given by $$\alpha(\triangle)={1\over2}(x_1y_2-x_2y_1)\ ,\tag{1}$$ as learned in high school analytic geometry. In the complex world we write $0$, $z_1$, $z_2$ for the three vertices, whereby the real coordinates $(x,y)$ are related to the corresponding $z$ via $$z:=x+iy,\quad x={z+\bar z\over 2},\quad y={z-\bar z\over 2i}\ .$$ Plugging this into $(1)$ we obtain $$\alpha(\triangle)={1\over8i}\bigl((z_1+\bar z_1)(z_2-\bar z_2)-(z_2+\bar z_2)(z_1-\bar z_1)\bigr)={1\over4i}(\bar z_1z_2-z_1\bar z_2)\ ,$$ which then can be rewritten as $$\alpha(\triangle)={1\over2}{\rm Im}(\bar z_1z_2)\ .$$ I leave it to you to get the formula for a triangle with vertices $z_0$, $z_1$, $z_2$ from this.

Answer 3

You can find the side lengths as $|z_1- z_2|, |z_2- z_3|, |z_3- z_1|$ and then use Heron's formula.

Answer 4

I assume you mean the oriented area of that triangle (since the number changes sign when you swap $z_2$ and $z_3$).

In order to verify it, allow me to write your formula as $$g(z_1,z_2,z_3)=\mathfrak{Im}\frac{\overline{z_1}(z_2-z_3)+z_3\overline{z_2}}{2}$$ It holds $$g(z_1+t,z_2+t,z_3+t)-g(z_1,z_2,z_3)= \mathfrak{Im}\frac{2\mathfrak{Re}(z_2\bar t)+\lvert t\rvert^2}{2}=0$$

So it is translation-invariant (hence, we need verify it only when $z_1=0$).

But that case is much easier.