Let z in mathbb{C} be such that frac{z^2+3i}{ | vedclass

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(B) Given the equation: $\frac{z^2+3i}{z-2+i} = 2+3i$Multiplying both sides by $(z-2+i)$,we get: $z^2+3i = (2+3i)(z-2+i)$$z^2+3i = 2z - 4 + 2i + 3iz -

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$-19-2i$

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(B) Given the equation: $\frac{z^2+3i}{z-2+i} = 2+3i$Multiplying both sides by $(z-2+i)$,we get: $z^2+3i = (2+3i)(z-2+i)$$z^2+3i = 2z - 4 + 2i + 3iz - 6i + 3i^2$Since $i^2 = -1$,$z^2+3i = 2z - 4 + 2i + 3iz - 6i - 3$$z^2+3i = z(2+3i) - 7 - 4i$$z^2 - z(2+3i) + 7 + 7i = 0$Let $z_1$ and $z_2$ be the roots of this quadratic equation. By Vieta's formulas,$z_1+z_2 = 2+3i$ and $z_1z_2 = 7+7i$.We need to find the sum of all possible values of $z^2$,which is $z_1^2 + z_2^2$.$z_1^2 + z_2^2 = (z_1+z_2)^2 - 2z_1z_2$$z_1^2 + z_2^2 = (2+3i)^2 - 2(7+7i)$$z_1^2 + z_2^2 = (4 + 9i^2 + 12i) - (14 + 14i)$$z_1^2 + z_2^2 = (4 - 9 + 12i) - 14 - 14i$$z_1^2 + z_2^2 = -5 + 12i - 14 - 14i$$z_1^2 + z_2^2 = -19 - 2i$

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

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