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If $x+i y=\frac{3}{2+\cos \theta+i \sin \theta}$, then $x^{2}+y^{2}$ is equal to
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Verified Answer
The correct answer is:
$4 x-3$
$x+i y$
$=\frac{3}{2+\cos \theta+i \sin \theta}=\frac{3(2+\cos \theta-i \sin \theta)}{(2+\cos \theta)^{2}+\sin ^{2} \theta}$
$\begin{array}{l}
=\frac{6+3 \cos \theta-3 i \sin \theta}{4+\cos ^{2} \theta+4 \cos \theta+\sin ^{2} \theta} \\
=\frac{6+3 \cos \theta-3 i \sin \theta}{5+4 \cos \theta} \\
=\left(\frac{6+3 \cos \theta}{5+4 \cos \theta}\right)+\left(\frac{-3 \sin \theta}{5+4 \cos \theta}\right)
\end{array}$
On equating real and imaginary parts, we get
$x=\frac{3(2+\cos \theta)}{5+4 \cos \theta}$
And $y=\frac{-3 \sin \theta}{5+4 \cos \theta}$
$\begin{array}{l}
\therefore x^{2}+y^{2}=\frac{9\left[4+\cos ^{2} \theta+4 \cos \theta+\sin ^{2} \theta\right]}{(5+4 \cos \theta)^{2}} \\
=\frac{9}{5+4 \cos \theta}=4\left(\frac{6+3 \cos \theta}{5+4 \cos ^{2} \theta}\right)-3 \\
=4 x-3
\end{array}$
$=\frac{3}{2+\cos \theta+i \sin \theta}=\frac{3(2+\cos \theta-i \sin \theta)}{(2+\cos \theta)^{2}+\sin ^{2} \theta}$
$\begin{array}{l}
=\frac{6+3 \cos \theta-3 i \sin \theta}{4+\cos ^{2} \theta+4 \cos \theta+\sin ^{2} \theta} \\
=\frac{6+3 \cos \theta-3 i \sin \theta}{5+4 \cos \theta} \\
=\left(\frac{6+3 \cos \theta}{5+4 \cos \theta}\right)+\left(\frac{-3 \sin \theta}{5+4 \cos \theta}\right)
\end{array}$
On equating real and imaginary parts, we get
$x=\frac{3(2+\cos \theta)}{5+4 \cos \theta}$
And $y=\frac{-3 \sin \theta}{5+4 \cos \theta}$
$\begin{array}{l}
\therefore x^{2}+y^{2}=\frac{9\left[4+\cos ^{2} \theta+4 \cos \theta+\sin ^{2} \theta\right]}{(5+4 \cos \theta)^{2}} \\
=\frac{9}{5+4 \cos \theta}=4\left(\frac{6+3 \cos \theta}{5+4 \cos ^{2} \theta}\right)-3 \\
=4 x-3
\end{array}$
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