Let, complex number $${\left( {\overline z } \right)^2} + {1 \over {{z^2}}}$$ is a new complex number $$w$$.

$$\therefore$$ $$w = {\left( {\overline z } \right)^2} + {1 \over {{z^2}}}$$

Now, let $$z = r{e^{i\theta }}$$ where $$r = |z|$$ and $$\theta$$ = argument

$$\therefore$$ $$w = {\left( {r{e^{ - i\theta }}} \right)^2} + {1 \over {{r^2}}}({e^{ - 2i\theta }})$$

$$ = \left( {{r^2} + {1 \over {{r^2}}}} \right){e^{ - 2i\theta }}$$

$$ = \left( {{r^2} + {1 \over {{r^2}}}} \right)(\cos 2\theta - i\sin 2\theta )$$

$$\therefore$$ Real part of $$w = {\mathop{\rm Re}\nolimits} (w) = \left( {{r^2} + {1 \over {{r^2}}}} \right)\cos 2\theta $$

Imaginary part of $$w = {\mathop{\rm Im}\nolimits} (w) = \left( {{r^2} + {1 \over {{r^2}}}} \right)( - \sin 2\theta )$$

Given both $${\mathop{\rm Re}\nolimits} (w)$$ and $${\mathop{\rm Im}\nolimits} (w)$$ are integer.

$$\therefore$$ Let $$Re(w) = {I_1}$$

and $${\mathop{\rm Im}\nolimits} (w) = {I_2}$$

$$\therefore$$ $$I_1^2 + I_2^2 = {\left( {{r^2} + {1 \over {{r^2}}}} \right)^2} = \alpha $$ (where $$\alpha$$ is an integer)

[Note : As I₁ and I₂ are integer so $$I_1^2 + I_2^2$$ is also an integer.]

Now, $${\left( {{r^2} + {1 \over {{r^2}}}} \right)^2} = \alpha $$

$$ \Rightarrow {r^4} + 2 + {1 \over {{r^4}}} = \alpha $$

$$ \Rightarrow {r^8} + 2{r^4} + 1 = {r^4}\alpha $$

$$ \Rightarrow {r^8} - (\alpha - 2){r^4} + 1 = 0$$

$$\therefore$$ $${r^4} = {{(\alpha - 2)\, \pm \,\sqrt {{{(\alpha - 2)}^2} - 4\,.\,1\,.\,1} } \over {2\,.\,1}}$$

$$ \Rightarrow r = {\left( {{{(\alpha - 2)\, \pm \,\sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$$

$$ \Rightarrow |z| = r = {\left( {{{(\alpha - 2)\, \pm \,\sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$$

In option only positive sign is given so ignoring negative sign we get,

$$|z| = r = {\left( {{{(\alpha - 2) + \sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$$ ....... (1)

From option (A),

$$|z| = {\left( {{{43 + 3\sqrt {205} } \over 2}} \right)^{{1 \over 4}}}$$

Comparing with (1), we get

$$\alpha - 2 = 43$$

$$ \Rightarrow \alpha = 45$$

Putting $$\alpha = 45$$ in (1), we get

$$|z| = {\left( {{{43 + \sqrt {45(41)} } \over 2}} \right)^{{1 \over 4}}}$$

$$ = {\left( {{{43 + \sqrt {9 \times 5 \times 41} } \over 2}} \right)^{{1 \over 4}}}$$

$$ = {\left( {{{43 + 3\sqrt {205} } \over 2}} \right)^{{1 \over 4}}}$$

$$\therefore$$ Option (A) is correct.

We can re-write

$$|z| = {\left( {{{(\alpha - 2) + \sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$$

as $$|z| = {\left( {{{2(\alpha - 2) + 2\sqrt {\alpha (\alpha - 4)} } \over 4}} \right)^{{1 \over 4}}}$$

Comparing with option (B) we get,

$$2(\alpha - 2) = 7$$

$$ \Rightarrow 2\alpha = 11$$

$$ \Rightarrow \alpha = {{11} \over 2}$$ $$\ne$$ Integer

$$\therefore$$ Option (B) is incorrect.

Similarly option (C) and (D) also incorrect.