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JEE Mains · Maths · STD 11 - 4.1 complex nubers

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

  1. A \(19-2 \mathrm{i}\)
  2. B \(-19-2 \mathrm{i}\)
  3. C \(19+2 i\)
  4. D \(-19+2 i\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-19-2 \mathrm{i}\)

Step-by-step Solution

Detailed explanation

\(z^2+3 i=z(2+3 i)-7-4 i\) \(\mathrm{z}^2-\mathrm{z}(2+3 \mathrm{i})+7+7 \mathrm{i}=0\) \(\begin{aligned} & z_1^2+z_2^2=\left(z_1+z_2\right)^2-2 z_1 z_2 \\ & =4-9+12 i-14-14 i \\ & =-19-2 i \end{aligned}\)
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