Show that Im(z_2bar z_1) = 2P , where P is the area of the…

Show that $\text{Im}(z_2\bar z_1) = 2P$, where $P$ is the area of the triangle whose vertices are $0, z_1, z_2$.

Step 1 · Set up

Let $z_1=x_1+iy_1$, $z_2=x_2+iy_2$, so $\bar z_1=x_1-iy_1$.

Step 2 · Compute $z_2\bar z_1$

$$ z_2\bar z_1 = (x_2+iy_2)(x_1-iy_1) = (x_1x_2+y_1y_2) + i(x_1y_2-x_2y_1) $$

so:

$$ \text{Im}(z_2\bar z_1) = x_1y_2-x_2y_1 $$

Step 3 · Triangle area formula

For a triangle with vertices $(0,0)$, $(x_1,y_1)$, $(x_2,y_2)$, the standard determinant/cross-product formula gives:

$$ P = \frac12\left|\det\begin{pmatrix}x_1&y_1\\x_2&y_2\end{pmatrix}\right| = \frac12|x_1y_2-x_2y_1| $$

(This follows from expanding the $3\times3$ vertex-determinant area formula along the row of zeros for the origin.)

Step 4 · Combine

Comparing Steps 2 and 3:

$$ |\text{Im}(z_2\bar z_1)| = |x_1y_2-x_2y_1| = 2P $$

Using the signed area $P_{\text{signed}}=\frac12(x_1y_2-x_2y_1)$ (positive for $0,z_1,z_2$ in counterclockwise order), the identity holds without absolute values:

$$ \boxed{\text{Im}(z_2\bar z_1) = 2P_{\text{signed}}} $$

Step 5 · Polar-form cross-check

Writing $z_1=|z_1|e^{i\theta_1}$, $z_2=|z_2|e^{i\theta_2}$: $z_2\bar z_1=|z_1||z_2|e^{i(\theta_2-\theta_1)}$, so:

$$ \text{Im}(z_2\bar z_1) = |z_1||z_2|\sin(\theta_2-\theta_1) $$

which matches the familiar triangle-area formula $P=\frac12|z_1||z_2|\sin\theta$, where $\theta=\theta_2-\theta_1$ is the angle between the two sides meeting at the origin.

Step 6 · Verify with examples

$z_1=1$, $z_2=i$: $z_2\bar z_1 = i\cdot1=i$, so $\text{Im}=1$. Triangle vertices $(0,0),(1,0),(0,1)$: right triangle, legs $1$, area $P=\frac12$. Check: $2P=1$. ✓

$z_1=2+i$, $z_2=1+3i$: $\bar z_1=2-i$, $z_2\bar z_1=(1+3i)(2-i)=2-i+6i-3i^2=5+5i$, so $\text{Im}=5$. Area: $P=\frac12|2\cdot3-1\cdot1|=\frac12|6-1|=\frac52$. Check: $2P=5$. ✓