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 ${z}^{2}$ is [JEE Main 2025, 3 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: b) $-19-2\mathrm{i}$. $\frac{z^2+3i}{z-2+i}=2+3i$ $z^2+3i=(z-2+i)(2+3i)$ $z^2+3i=2z-4+2i+3iz-6i-3$ $z^2+3i=(2z-7)+i(3z-4)$ $z^2-(2+3i)z+(7+7i)=0$ This is a quadratic in $z$. $z_1+z_2=2+3i$ $z_1z_2=7+7i$ $z_1^2+z_2^2=(z_1+z_2)^2-2z_1z_2$ $=(2+3i)^2-2(7+7i)$ $=4-9+12i-14-14i$ $=-19-2i$

Let \(\mathrm{z}\in \mathrm{C}\) be such that \(\frac{{\mathrm{z}}^{2}+3\mathrm{i}}{\mathrm{z}-2+\mathrm{i}}=2+3\mathrm{…

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