Q. If $z = x + iy$, then area of the triangle whose vertices are points $z,\ iz,\ z + iz$ is
A) $\frac{1}{2}|z|^{2}$
B) $\frac{1}{4}|z|^{2}$
C) $|z|^{2}$
D) $\frac{3}{2}|z|^{2}$
Explanation
(a)
The vertices of the triangle are $z,\ iz,\ z + iz$
or $x + iy,\ - y + ix,\ \ (x - y) + i(x + y)$
$\therefore$ Required area$= \frac{1}{2}\left| \begin{matrix}
x & y & 1 \\
- y & x & 1 \\
x - y & x + y & 1
\end{matrix} \right|$
$= \frac{1}{2}|\lbrack x(x - x - y) - y( - y - x + y) + 1( - yx - y^{2} - x^{2} + xy)\rbrack|$
$= \frac{1}{2}\left( x^{2} + y^{2} \right) = \frac{1}{2}|z|^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.