The area of the triangle whose vertices are complex numbers $$z, iz, z + iz$$ in the Argand diagram is

Let the complex number be written in rectangular form as $$z = x + iy$$, where $$x, y \in \mathbb{R}$$.

On the Argand diagram the three vertices are
  $$A \equiv z = (x,\,y)$$,
  $$B \equiv iz = i(x+iy) = -y + ix = (-y,\,x)$$,
  $$C \equiv z + iz = (x + iy) + (-y + ix) = (x - y) + i(x + y) = (x-y,\,x+y).$$

For any triangle with vertices $$A(x_1,y_1),\,B(x_2,y_2),\,C(x_3,y_3)$$, the area is

$$\text{Area} = \frac12 \left| \begin{vmatrix} x_2-x_1 & y_2-y_1\\ x_3-x_1 & y_3-y_1 \end{vmatrix} \right|.$$

Compute the two side vectors using the coordinates found above:
$$\overrightarrow{AB} = (x_2-x_1,\,y_2-y_1) = (-y-x,\,x-y),$$
$$\overrightarrow{AC} = (x_3-x_1,\,y_3-y_1) = (-y,\,x).$$

The determinant of these two vectors is
$$\begin{vmatrix} -\,y - x & x - y\\ -\,y & x \end{vmatrix} = (-y - x)\,x - (x - y)(-y) = -xy - x^{2} + xy - y^{2} = -\,(x^{2} + y^{2}).$$

Taking absolute value and then halving, the area becomes
$$\text{Area} = \frac12 \left| -\,(x^{2} + y^{2}) \right| = \frac12\,(x^{2} + y^{2}) = \frac12\,|z|^{2},$$
because $$|z|^{2} = x^{2} + y^{2}.$$

Therefore the required area is $$\dfrac12\,|z|^{2}$$.

Option B which is: $$\dfrac{1}{2}|z|^{2}$$