Explanation

(b)
Vertices of the triangle are $0 = 0 + i0,\ \mathcal{z =}x + iy$
and $\mathcal{z}e^{i\alpha} = (x + iy)(\cos{\alpha + i\sin{\alpha)}}$
$= (x\cos{\alpha - y\sin{\alpha) + i\ (y\cos{\alpha + x\sin{\alpha)}}}}$
$\therefore$ Area of triangle
$= \frac{1}{2}\ \left| \begin{matrix}
0 & 0 & 1 \\
x & y & 1 \\
(x\cos{\ \alpha - y\ \sin\ \alpha)} & (y\cos{\ \alpha + x\ \sin\ \alpha)} & 1
\end{matrix} \right|$
$= \frac{1}{2}\lbrack xy\cos{\alpha + x^{2}\sin{\alpha - xy\cos{\alpha + y^{2}\sin{\alpha\rbrack\ }}}}$
$= \frac{1}{2}\left( x^{2} + y^{2} \right)\sin{\alpha = \frac{1}{2}|}\mathcal{z}\left. \ \right|^{2}\sin{\alpha\ \ \ \ \ \ \ \ (\because\left| \mathcal{z} \right| = \sqrt{x^{2} + y^{2}})}$