Let z C be such that (z^2+3 i)/(z-2+i)=2+3 i.

Question

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

Accepted Answer

$\begin{gathered} \frac{{z}^{2}+3i}{z-2+i}=2+3i \ {z}^{2}+3i=(z-2+i)(2+3i) \ {z}^{2}+3i=2z-4+2i+3iz-6i-3 \ {z}^{2}+3i=(2z-7)+i(3z-4) \ {z}^{2}-(2+3i)z+(7+7i)=0 \ z \ {z}_{1}+{z}_{2}=2+3i \ {z}_{1}+{z}_{2}=7+7i \ {z}_{2}^{1}+{z}_{2}^{2}={({z}_{1}+{z}_{2})}^{2}-2{z}_{1}{z}_{2} \ =(2+3i{)}^{2}-2(7+7i) \ =4-9+12i-14-14i \ =-19-2i \end{gathered}$

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