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Q.
The area of the triangle on the complex plane formed by the complex numbers $z, i z$ and $z+i z$, is
Complex Numbers and Quadratic Equations
Solution:
Let $z=x+i y$. Then, $i z=-y+i x$. Therefore,
$z+i z=(x-y)+i(x+y)$
Required area of triangle
$ =\text { Area of triangle formed by }(x, y)(-y, x) \text { and }(x-y, x+y) $
$ =\frac{1}{2}|x(x-x-y)-y(x+y-y)+(x-y)(y-x)| $
$ =\frac{1}{2}\left|2 x y-2 x y-x^2-y^2\right|$
$ =\frac{1}{2}\left(x^2+y^2\right)=\frac{|z|^2}{2}$
Geometrically
Area of $\triangle A B C-\frac{1}{2}$ area of square $O \Lambda C B$
$=\frac{1}{2}|z|^2$
