Mod of the cubics!

Denoting $j$ as the imaginary unit:

$\large \displaystyle z=re^{j\theta}$

Then:

$\large \displaystyle |z| = |z^2| + |z^3|$

$\large \displaystyle r = r^2 + r^3$

Since $r \neq 0$ :

$\large \displaystyle r^2 + r - 1 =0$

Since $r > 0$ :

$\color{#20A900} \boxed{\large \displaystyle r = \frac{\sqrt{5}-1}{2}}$

Thus:

$\large \displaystyle a^2 + b^2 = r^2$

$\large \displaystyle a^2 + b^2 = \frac{6 - 2\sqrt{5}}{4}$

$\large \displaystyle b^2 = \frac{6 - 2\sqrt{5}-4a^2}{4}$

$\color{#20A900} \boxed{\large \displaystyle |b| = \sqrt{\frac{-2a^2 - \sqrt{5} +3}{2}}}$

Thus:

$\color{#3D99F6} \large \displaystyle A=2, B=5, C=3,D=2, \boxed{\large \displaystyle A+B+C+D=12}$

By the properties of modulus, $|z|=|{ z }^{ 2 }|+|{ z }^{ 3 }|$ where $z = a + bi$ rewrites as $|z|=|{ z }|^2+|{ z }|^3$ .

Thus solving for $|z|$ , we have $|z| = 0$ , which is rejected because $a>0 \Rightarrow |z| > 0$ , or $|z| = \frac{-1\pm \sqrt{5}}{2}$ . Again, $|z| > 0$ rejects $\frac{-1\ - \sqrt{5}}{2}$ as a possible solution, leaving the correct answer

$|z| = \frac{\sqrt{5}-1}{2}$ $(|z|^2) = \frac{1}{2}(3 - \sqrt{5}) = a^2 + b^2$ $b^2 = \frac{1}{2}((3) - \sqrt{5}) - a^2 = \frac{-2a^2 - \sqrt{5} + 3}{2}$ $\sqrt{b^2} = |b| = \sqrt{\frac{-2a^2 - \sqrt{5} + 3}{2}}.$

$A = 2, B = 5, C = 3,$ and $D = 2$ , thus $A + B + C + D = \boxed{12}$ .