$$\begin{aligned} &\begin{aligned} & \frac{z^2+3 i}{z-2+i}=2+3 i \\ & z^2+3 i=(z-2+i)(2+3 i) \\ & z^2+3 i=2 z-4+2 i+3 i z-6 i-3 \\ & z^2+3 i=(2 z-7)+i(3 z-4) \\ & z^2-(2+3 i) z+(7+7 i)=0 \end{aligned}\\ &\text { This is a quadratic in } z \text {. }\\ &\begin{aligned} & z_1+z_2=2+3 i \\ & z_1+z_2=7+7 i \\ & z_1^2+z_2^2=\left(z_1+z_2\right)^2-2 z_1 z_2 \\ & =(2+3 i)^2-2(7+7 i) \\ & =4-9+12 i-14-14 i \\ & =-19-2 i \end{aligned} \end{aligned}$$