JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(\quad S=\left\{z \in C-\{i, 2 i\}: \frac{z^2+8 i z-15}{z^2-3 i z-2} \in R \right\}\). \(\alpha-\frac{13}{11} i \in S , \alpha \in R -\{0\}\), then \(242 \alpha^2\) is equal to
- A \(1680\)
- B \(1681\)
- C \(1682\)
- D \(1683\)
Answer & Solution
Correct Answer
(A) \(1680\)
Step-by-step Solution
Detailed explanation
\(\left(\frac{z^2+81 z-15}{z^2-3 i z-2}\right) \in R\) \(\Rightarrow 1+\frac{(11 i z-13)}{\left(z^2-3 i z-2\right)} \in R\) Put \(z=\alpha-\frac{13}{11} i\) \(\Rightarrow\left(z^2-3 i z-2\right)\) is imaginary Put \(z=x+i y\)…
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