Let A = {θ ∈ [0, 2π] : 1+10 Re} (2 cosθ + i sinθ/cosθ - 3 isinθ) = θ}. Then ∑ θ∈A θ^2 is equal to

Question

Let \(\mathrm{A}=\left\{\theta \in[0,2 \pi]: 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}\right)=0\right\}.\) Then \(\sum\limits_{\theta \in \mathrm{A}} \theta^{2}\) is equal to

(1) \(\frac{21}{4} \pi^{2}\)

(2) \(8 \pi^{2}\)

(3) \(\frac{27}{4} \pi^{2}\)

(4) \(6 \pi^{2}\) 

Answer 1

Correct option is: (1) \(\frac{21}{4} \pi^{2}\)  

\(1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0\)

\(\therefore \mathrm{z}+\overline{\mathrm{z}}=2 \operatorname{Re}(\mathrm{z})\)

\(\frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}+\frac{2 \cos \theta-\mathrm{i} \sin \theta}{\cos \theta+3 \mathrm{i} \sin \theta}=2 \times\left(\frac{-1}{10}\right)\)

\(\frac{\left(2 \cos ^{2} \theta-3 \sin ^{2} \theta\right)+\left(2 \cos ^{2} \theta\right)-\left(3 \sin ^{2} \theta\right)}{\cos ^{2} \theta+9 \sin ^{2} \theta}=\frac{-2}{10}\)

\(\Rightarrow \frac{2 \cos ^{2} \theta-3 \sin ^{2} \theta}{\cos ^{2} \theta+9 \sin ^{2} \theta}=\frac{-1}{10}\)

\(\Rightarrow 20 \cos ^{2} \theta-30 \sin ^{2} \theta=-\cos ^{2} \theta-9 \sin ^{2} \theta\)

\(21 \cos ^{2} \theta-21 \sin ^{2} \theta=0\)

\(\Rightarrow \cos 2 \theta=0\)

\(2 \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}\)

\(\Rightarrow \sum \theta^{2}=\frac{\pi^{2}}{16}+\frac{9 \pi^{2}}{16}+\frac{25 \pi^{2}}{16}+\frac{49 \pi^{2}}{16}=\frac{84 \pi^{2}}{16}=\frac{21 \pi^{2}}{4}\)