Q64 / 120 · VITEEE 2014 · Mathematics
If $x+i y=\frac{3}{2+\cos \theta+i \sin \theta}$, then $x^{2}+y^{2}$ is equal to
- A$3 x-4$
- Correct answerB$4 x-3$
- C$4 x+3$
- DNone of these
Solution
$x+i y$
$=\frac{3}{2+\cos \theta+i \sin \theta}=\frac{3(2+\cos \theta-i \sin \theta)}{(2+\cos \theta)^{2}+\sin ^{2} \theta}$
$\begin{array}{l}
=\frac{6+3 \cos \theta-3 i \sin \theta}{4+\cos ^{2} \theta+4 \cos \theta+\sin ^{2} \theta} \\
=\frac{6+3 \cos \theta-3 i \sin \theta}{5+4 \cos \theta} \\
=\left(\frac{6+3 \cos \theta}{5+4 \cos \theta}\right)+\left(\frac{-3 \sin \theta}{5+4 \cos \theta}\right)
\end{array}$
On equating real and imaginary parts, we get
$x=\frac{3(2+\cos \theta)}{5+4 \cos \theta}$
And $y=\frac{-3 \sin \theta}{5+4 \cos \theta}$
$\begin{array}{l}
\therefore x^{2}+y^{2}=\frac{9\left[4+\cos ^{2} \theta+4 \cos \theta+\sin ^{2} \theta\right]}{(5+4 \cos \theta)^{2}} \\
=\frac{9}{5+4 \cos \theta}=4\left(\frac{6+3 \cos \theta}{5+4 \cos ^{2} \theta}\right)-3 \\
=4 x-3
\end{array}$
Hence, the correct answer is option: 'B'.