[UPLIFTJEE]

Let $\mathrm{z} \in \mathrm{C}$ be such that $\frac{\mathrm{z}^{2}+3 \mathrm{i}}{\mathrm{z}-2+\mathrm{i}}=2+3 \mathrm{i}$. Then the sum of all possible values of $\mathrm{z}^{2}$ is
[JEE Main 03 April 2025 S I]
(A) 19 – 2i
(B) 19 + 2i
(C) $-19-2 \mathrm{i}$
(D) $-19+2 \mathrm{i}$

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[UPLIFTJEE]

Solution

$\frac{z^{2}+3 i}{z-2+i}=2+3 i \Rightarrow z^{2}+3 i=z(2+3 i)-7-4 i \Rightarrow z^{2}-z(2+3 i)+7+7 i=0$ has roots $z_{1}$ and $z_{2}$

\[ \mathrm{z}_{1}+\mathrm{z}_{2}=2+3 \mathrm{i}, \mathrm{z}_{1} \mathrm{z}_{2}=7+7 \mathrm{i} \]

Required sum is $\mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}=\left(\mathrm{z}_{1}+\mathrm{z}_{2}\right)^{2}-2 \mathrm{z}_{1} \mathrm{z}_{2}=(2+3 \mathrm{i})^{2}-2(7+7 \mathrm{i})=4-9+12 \mathrm{i}-14-14 \mathrm{i}=-19-2 \mathrm{i}$

[UPLIFTJEE]

Answer

C

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