Explanation

(b)
$x + iy = \frac{3}{2 + \cos{\theta + i\sin\theta}}$
$= \ \frac{3(2 + \cos{\theta - i\sin{\theta)}}}{\left( 2 + \cos\theta \right)^{2} + \sin^{2}\theta}$
$= \frac{6 + 3\cos{\theta - 3i\sin\theta}}{4 + \cos^{2}{\theta + 4\cos{\theta + \sin^{2}\theta}}}$
$= \left\lbrack \frac{6 + 3\cos\theta}{5 + 4\cos\theta} \right\rbrack + i\left\lbrack \frac{- 3\sin\theta}{5 + 4\cos\theta} \right\rbrack$
On equating the real and imaginary parts on both sides, we get
$x = \frac{3(2 + \cos{\theta)}}{(5 + 4\cos{\theta)}},\ \ y = \frac{- 3\sin\theta}{5 + 4\cos\theta}$
$\therefore\ \ x^{2} + y^{2} = \frac{9}{\left( 5 + 4\cos\theta \right)^{2}}\lbrack 4 + \cos^{2}{\theta + 4\cos{\theta + \sin^{2}{\theta\rbrack\ }}}$
$= \ \frac{9}{5 + 4\cos\theta} = 4\left\lbrack \frac{6 + 3\cos\theta}{5 + 4\cos\theta} \right\rbrack - 3$
$= 4x - 3$