Let $$z \in \mathbb{C}$$ be such that $$\frac{z^2 + 3i}{z - 2 + i} = 2 + 3i$$. Then the sum of all possible values of $$z^2$$ is

The given condition is
$$\frac{z^{2}+3i}{z-2+i}=2+3i$$.

First cross-multiply:
$$z^{2}+3i=(2+3i)\,(z-2+i).$$

Expand the right-hand side.
We need $$(2+3i)(z-2+i)=(2+3i)z+(2+3i)(-2+i).$$

Compute the constant product:
$$(2+3i)(-2+i)=-4-6i+2i+3i^{2}=-4-4i-3=-7-4i.$$

Hence
$$z^{2}+3i=(2+3i)z-7-4i.$$

Bring all terms to the left:
$$z^{2}-(2+3i)z+(3i+7+4i)=0.$$

Simplify the constant term $$3i+4i=7i$$, so the quadratic in $$z$$ is
$$z^{2}-(2+3i)z+(7+7i)=0.$$

For a quadratic $$z^{2}+Bz+C=0$$ with roots $$z_{1},z_{2}$$:
• Sum of roots $$S=z_{1}+z_{2}=-B.$$
• Product of roots $$P=z_{1}z_{2}=C.$$

Here $$B=-(2+3i)$$ and $$C=7+7i$$, hence
$$S=2+3i,\qquad P=7+7i.$$

We want the sum of all possible values of $$z^{2}$$, namely $$z_{1}^{2}+z_{2}^{2}$$.
Use the identity
$$z_{1}^{2}+z_{2}^{2}=(z_{1}+z_{2})^{2}-2z_{1}z_{2}=S^{2}-2P.$$

Compute $$S^{2}$$:
$$(2+3i)^{2}=4+12i+9i^{2}=4+12i-9=-5+12i.$$

Compute $$2P$$:
$$2P=2(7+7i)=14+14i.$$

Therefore
$$z_{1}^{2}+z_{2}^{2}=(-5+12i)-(14+14i)=-19-2i.$$

Thus the sum of all possible values of $$z^{2}$$ is $$-19-2i$$, which matches Option B.