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

2. Convert each vertex into Cartesian coordinates.

   • $A(z) \Rightarrow A(x,y)$
   • $B(iz)$: $iz = i(x+iy) = ix - y = -y + ix$ so $B(-y,x)$
   • $C(z+iz)$: $z+iz = (x+iy)+(-y+ix) = (x-y) + i(x+y)$ so $C(x-y, x+y)$

3. Observe that $C = A + B$ in vector form, since $(x,y) + (-y,x) = (x-y,x+y).$

   Hence triangle $ABC$ has sides from the origin-like vector relation, and its area is $\text{Area} = \frac12 \left| \det\begin{pmatrix}x & y\\ -y & x\end{pmatrix} \right|.$

4. Compute the determinant: $\det\begin{pmatrix}x & y\\ -y & x\end{pmatrix} = x^2 + y^2.$

   Therefore, $\text{Area} = \frac12 (x^2+y^2).$

5. Since $|z|^2 = x^2+y^2,$ we get $\boxed{\text{Area} = \frac12 |z|^2}.$

6. Check options:

   • A: $1$ ❌
   • B: $\frac12 |z|^2$ ✅
   • C: $\frac12 |z+iz|^2$; note $|z+iz|^2 = |z(1+i)|^2 = 2|z|^2$, so this equals $|z|^2$, not the area ❌
   • D: $\frac12$ ❌

Therefore the correct option is B.