JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(\mathrm{A}=\)\(\left\{\theta \in[0,2 \pi]: 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0\right\} .\)
Then \(\sum_{\theta \in A} \theta^2\) is equal to
- A \(\frac{21}{4} \pi^2\)
- B \(8 \pi^2\)
- C \(\frac{27}{4} \pi^2\)
- D \(6 \pi^2\)
Answer & Solution
Correct Answer
(A) \(\frac{21}{4} \pi^2\)
Step-by-step Solution
Detailed explanation
\(1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0 \) \( \therefore \mathrm{z}+\overline{\mathrm{z}}=2 \operatorname{Re}(\mathrm{z}) \)…
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