The area of the triangle whose vertices are complex numbers $z, i z, z+i z$ in the Argand diagram is
The area of the triangle whose vertices are complex numbers $z, i z, z+i z$ in the Argand diagram is
$2|z|^2$
$1 / 2|z|^2$
$4|z|^2$
$|z|^2$
Solution
Vertices of triangle in complex form is $z, i z, z+i z$
In cartesian form vertices are $(x, y),(-y, x)$ and $(x-y, x+y)$
$
\begin{aligned}
& \therefore \text { Area of triangle }=\frac{1}{2}\left|\begin{array}{ccc}
x & y & 1 \\
-y & x & 1 \\
x-y & x+y & 1
\end{array}\right| \\
& =\frac{1}{2}[x(x-x-y)-y(-y-x+y)+1 \\
& \left.\left(-y x-y^2-x^2+x y\right)\right]
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{2}\left[-x y+x y-y^2-x^2\right]=\frac{1}{2}\left(x^2+y^2\right) \\
& (\because \text { Area can not be negative }) \\
& =\frac{1}{2}|z|^2 \quad\left(\because z=x+i y,|z|^2=x^2+y^2\right)
\end{aligned}
$