Triangle in complex plane

Question

Let T be a triangle in complex plane with vertices at $0,z_1, z_2$. Show that the area of T is equal to $\frac12|\mathsf{Im}(z_1\overline{z_2})|$ any ideas on this?

Answer 1

Hint:  write it as $\;\dfrac{1}{2} \cdot |z_1|\cdot|z_2| \cdot \left| \mathsf{Im}\left(\dfrac{\;\;\;\dfrac{z_1}{|z_1|}\;\;\;}{\dfrac{z_2}{|z_2|}}\right) \right|\;$ and remember that the area of a triangle equals half the product of two sides times the $\sin$ of the angle between them.

Answer 2

Let $Z_1=(x_1,y_1)$ and $Z_2=(x_2,y_2)$ .

Note that $$ \frac12|\mathsf{Im}(z_1\overline{z_2})| = \frac {1}{2} |x_2y_1 -x_1y_2|$$

The area of the triangle made by vector $$ V_1= (x_1, y_1,0)$$ and $$ V_2= (x_2, y_2,0)$$ is $1/2$ times the norm of the cross product of $V_1$ adn $V_2$.

Note that $$ ||V_1 \times V_2|| = |x_2y_1 -x_1y_2|$$.

Thus the desired formula for the area of the triangle has been proved.

Answer 3

Actually, I disagree with your initial assertion that

$$A=\frac12|\mathsf{Im}(z_1\overline{z_2})|$$

I believe that

$$A=\frac12\mathsf{Im}(\overline{z_1}z_2)$$

The difference is only in the sign of the area. But with area in the complex plane having a sign that is dependent upon the trajectory of the curve, i.e,. clockwise versus anti-clockwise, there is a difference.

I have verified my result by direct comparison with numerical integration of the area in the complex plane with

$$A=\frac12\int \mathsf{Im}(\overline{z}\dot z)~du$$

using random values of $z_{1,2}$ about the plane.

Answer 4

The area of a triangle is $\frac{1}{2}ab\sin\gamma,$ where $a,b$ are sides and $\gamma$ is the angle intercepted by these sides.

Use $z_1=ae^{i\alpha}, z_2=be^{i\beta}.$ Then $z_1\overline{z_2}=abe^{i(\beta-\alpha)},$ from where $$\frac{1}{2}Im(z_1\overline{z_2})=\frac{1}{2}ab\sin(\beta-\alpha).$$ Since the angle in the triangle is $(\beta-\alpha)$ or $(2\pi-(\beta-\alpha)),$ the area is the absolute value of the above expression.