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Area of the triangle formed by the complex numbers $\mathrm{z}, \mathrm{iz}$ and $\mathrm{z}+\mathrm{iz}$ in the Argand diagram as vertices is
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Verified Answer
The correct answer is:
$\frac{1}{2} \cdot|z|^2$
Let $Z=x+i y ; i z=-y+i x$
and $\mathrm{Z}+\mathrm{iz}=(\mathrm{x}-\mathrm{y})+\mathrm{i}(\mathrm{x}+\mathrm{y})$
$\therefore$ Area of triangle
$\begin{aligned} & \varnothing=\frac{1}{2}\left|\begin{array}{ccc}x & y & 1 \\ x-y & x+y & 1 \\ -y & x & 1\end{array}\right| \\ & =\frac{1}{2}\left(x^2+y^2\right)=\frac{1}{2}|z|^2\end{aligned}$
and $\mathrm{Z}+\mathrm{iz}=(\mathrm{x}-\mathrm{y})+\mathrm{i}(\mathrm{x}+\mathrm{y})$
$\therefore$ Area of triangle
$\begin{aligned} & \varnothing=\frac{1}{2}\left|\begin{array}{ccc}x & y & 1 \\ x-y & x+y & 1 \\ -y & x & 1\end{array}\right| \\ & =\frac{1}{2}\left(x^2+y^2\right)=\frac{1}{2}|z|^2\end{aligned}$
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