Let ¯ z denote the complex conjugate of a complex number z . If z...

Let $z = r . {e}^{i\theta}$

So, ${\left(\overset{¯}{z}\right)}^{2} + \dfrac{1}{{z}^{2}} = \left({r}^{2} + \dfrac{1}{{r}^{2}}\right){e}^{- 2i\theta} = a + ib$ (say), given $a, b \in Z$

Now taking mod both side and squaring we get,

${\left({r}^{2} + \dfrac{1}{{r}^{2}}\right)}^{2} = {a}^{2} + {b}^{2}$

$⇒ {r}^{8} - \left({a}^{2} + {b}^{2} - 2\right){r}^{4} + 1 = 0$

$⇒ {r}^{4} = \dfrac{\left({a}^{2} + {b}^{2} - 2\right) \pm \sqrt{{\left({a}^{2} + {b}^{2} - 2\right)}^{2} - 4}}{2}$

Now comparing from option we get, for ${a}^{2} + {b}^{2} = 45$ (i.e \left(a, b\right) = \left(\pm 6, \pm 3\right)$or \left(\pm 3, \pm 6\right)$

We get $r = \left(\dfrac{43 + 3\sqrt{205}}{2}\right)\dfrac{1}{4}$