Explanation

(b)
Given, $\frac{3}{2 + \cos{\theta + i\sin\theta}} = a + ib$
$\Rightarrow \ \ \ \frac{3\lbrack\left( 2 + \cos\theta \right) - i\ sin\ \theta\rbrack}{\left( 2 + \cos\theta \right)^{2} + \sin^{2}\theta} = a + ib$
$\Rightarrow \frac{3\lbrack 2 + \cos{\theta - i\sin\theta}\rbrack}{5 + 4\cos\theta} = a + ib$
$\Rightarrow \ \ \ \ \ a = \frac{3(2 + \cos\theta)}{5 + 4\cos\theta},\ \ b = - \frac{3\sin\theta}{5 + 4\cos\theta}$
$\therefore\ \ \ \ (a - 2)^{2} + b^{2} = \left( \frac{6 + 3\cos\theta}{5 + 4\cos\theta} - 2 \right)^{2} + \frac{9\sin^{2}\theta}{\left( 5 + 4\cos\theta \right)^{2}}$
$= \frac{\left( - 4 - 5\cos\theta \right)^{2} + 9\sin^{2}\theta}{\left( 5 + 4\cos\theta \right)^{2}}$
$= \frac{16 + 25\cos^{2}{\theta + 40\cos{\theta + 9\sin^{2}\theta}}}{\left( 5 + 4\cos\theta \right)^{2}}$
$= \frac{16 + 16\cos^{2}{\theta + 40\cos{\theta + 9}}}{\left( 5 + 4\cos\theta \right)^{2}}$
$= \frac{\left( 5 + 4\cos\theta \right)^{2}}{\left( 5 + 4\cos\theta \right)^{2}} = 1$