Let A=[0,2]: 1+10Re(2+i-3i)=0.

Question

Let $A=\left\{	heta\in[0,2\pi]: 1+10\operatorname{Re}\left(\dfrac{2\cos	heta+i\sin	heta}{\cos	heta-3i\sin	heta}ight)=0ight\}$. Then $\displaystyle\sum_{	heta\in A}	heta^2$ is equal to:

Accepted Answer

\[\begin{aligned}
& 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0 \
& \therefore \mathrm{z}+\overline{\mathrm{z}}=2 \operatorname{Re}(\mathrm{z}) \
& \frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}+\frac{2 \cos \theta-\mathrm{i} \sin \theta}{\cos \theta+3 \mathrm{i} \sin \theta}=2 \times\left(\frac{-1}{10}\right) \
& \frac{\left(2 \cos ^2 \theta-3 \sin ^2 \theta\right)+\left(2 \cos ^2 \theta\right)-\left(3 \sin ^2 \theta\right)}{\cos ^2 \theta+9 \sin ^2 \theta}=\frac{-2}{10} \
& \Rightarrow \frac{2 \cos ^2 \theta-3 \sin ^2 \theta}{\cos ^2 \theta+9 \sin ^2 \theta}=\frac{-1}{10} \
& \Rightarrow 20 \cos ^2 \theta-30 \sin ^2 \theta=-\cos ^2 \theta-9 \sin ^2 \theta \
& 21 \cos ^2 \theta-21 \sin ^2 \theta=0 \
& \Rightarrow \cos 2 \theta=0 \
& 2 \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \
& \Rightarrow \sum \theta^2=\frac{\pi^2}{16}+\frac{9 \pi^2}{16}+\frac{25 \pi^2}{16}+\frac{49 \pi^2}{16}=\frac{84 \pi^2}{16}=\frac{21 \pi^2}{4}
\end{aligned}\]

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