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

Question

Let $A={	heta \in [0,2\pi ]:1+10Re(\frac{2\cos 	heta +i\sin 	heta }{\cos 	heta -3i\sin 	heta })=0}$. Then $\sum _{	heta \in A}{	heta }^{2}$ is equal to

Accepted Answer

$\begin{gathered} 1+10Re(\frac{2\cos 	heta +i\sin 	heta }{\cos 	heta -3i\sin 	heta })=0 \ 	herefore z+\vec{z}=2Re(z) \ \frac{2\cos 	heta +i\sin 	heta }{\cos 	heta -3i\sin 	heta }+\frac{2\cos 	heta -i\sin 	heta }{\cos 	heta +3i\sin 	heta }=2	imes (\frac{-1}{10}) \ \frac{(2{\cos }^{2}	heta -3{\sin }^{2}	heta )+(2{\cos }^{2}	heta )-(3{\sin }^{2}	heta )}{{\cos }^{2}	heta +9{\sin }^{2}	heta }=\frac{-2}{10} \ \Rightarrow \frac{2{\cos }^{2}	heta -3{\sin }^{2}	heta }{{\cos }^{2}	heta +9{\sin }^{2}	heta }=\frac{-1}{10} \ \Rightarrow 20{\cos }^{2}	heta -30{\sin }^{2}	heta =-{\cos }^{2}	heta -9{\sin }^{2}	heta \ 21{\cos }^{2}	heta -21{\sin }^{2}	heta =0 \ \Rightarrow \cos 2	heta =0 \ 2	heta =\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2} \ \Rightarrow \sum {	heta }^{2}=\frac{{\pi }^{2}}{16}+\frac{9{\pi }^{2}}{16}+\frac{25{\pi }^{2}}{16}+\frac{49{\pi }^{2}}{16}=\frac{84{\pi }^{2}}{16}=\frac{21{\pi }^{2}}{4} \end{gathered}$

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