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 :
Solution
<p>$$\begin{aligned}
&\begin{aligned}
& \frac{z^2+3 i}{z-2+i}=2+3 i \\
& z^2+3 i=(z-2+i)(2+3 i) \\
& z^2+3 i=2 z-4+2 i+3 i z-6 i-3 \\
& z^2+3 i=(2 z-7)+i(3 z-4) \\
& z^2-(2+3 i) z+(7+7 i)=0
\end{aligned}\\
&\text { This is a quadratic in } z \text {. }\\
&\begin{aligned}
& z_1+z_2=2+3 i \\
& z_1+z_2=7+7 i \\
& z_1^2+z_2^2=\left(z_1+z_2\right)^2-2 z_1 z_2 \\
& =(2+3 i)^2-2(7+7 i) \\
& =4-9+12 i-14-14 i \\
& =-19-2 i
\end{aligned}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
This question is part of PrepWiser's free JEE Main question bank. 223 more solved questions on Complex Numbers and Quadratic Equations are available — start with the harder ones if your accuracy is >70%.