Explanation

(c)
Let $\mathcal{z =}r\ (\cos{\theta + i\sin{\theta)}}$
Given that $\left| \mathcal{z +}\frac{1}{\mathcal{z}} \right| = a\ \Rightarrow \left| \mathcal{z +}\frac{1}{\mathcal{z}} \right|^{2} = a^{2}$
$\Rightarrow r^{2} + \frac{1}{r^{2}} + 2\cos 2\theta = a^{2}$ …(i)
On differentiating w.r.t. $\theta,$ we get
$2r\frac{dr}{d\theta} - \frac{2}{r^{3}}\frac{dr}{d\theta} - 4\sin{2\theta = 0}$
$\Rightarrow \frac{dr}{d\theta}\left( 2r - \frac{2}{r^{3}} \right) = 4\sin{2\theta}$
For maximum or minimum, put $\frac{dr}{d\theta} = 0,$ we get
$\theta = 0,\frac{\pi}{2}$
$\therefore r$ is maximum for $\theta = \frac{\pi}{2},$ therefore from Eq.(i)
$r^{2} + \frac{1}{r^{2}} - 2 = a^{2}\ \Rightarrow \ r - \frac{1}{r} = a$
$\Rightarrow \ r^{2} - ar - 1 = 0$
$\Rightarrow \ r = \frac{a + \sqrt{a^{2} + 4}}{2}$