Let $\overline{z}$ denote the complex conjugate of a complex number  $z$. If $z$ is a non-zero complex number for which both real and imaginary parts of  ${\left(\overline{z}\right)}^2+\frac{1}{z^2}$  are integers, then which of the following is/are possible value(s) of  $\left|z\right|$ ?

$$\begin{gathered} z=x+iy \\ {\left(\overline{z}\right)}^2+\frac{1}{z^2}=a+ib,a,b\in I \\ \Rightarrow {\left(\overline{z}\right)}^2+\frac{{\left(\overline{z}\right)}^2}{z^2{\overline{z}}^2}=a+ib,a,b\in I \end{gathered}$$

$\Rightarrow {\overline{z}}^2(1+\frac{1}{{\left|z\right|}^4})=a+ib$ [${\left|z\right|}^4\to$ real number]

$\{x^2-y^2-i\left(2xy\right)\}(1+\frac{1}{{\left|z\right|}^4})=a+ib$

On comparing,
$(x^2-y^2)(1+\frac{1}{{\left|z\right|}^4})=a$………(i)
 and  $-2xy(1+\frac{1}{{\left|z\right|}^4})=b$ ………(ii)
on squaring and adding eqs. (i)  and (ii). We get
$$\begin{gathered} (x^4+y^4-2x^2{y}^2+4x^2{y}^2){(1+\frac{1}{{\left|z\right|}^4})}^2=a^2+b^2 \\ {(x^2+y^2)}^2{(1+\frac{1}{{\left|z\right|}^4})}^2=a^2+b^2 \\ {\left|z\right|}^4{(1+\frac{1}{{\left|z\right|}^4})}^2=a^2+b^2 \\ {({\left|z\right|}^2+\frac{1}{{\left|z\right|}^2})}^2=a^2+b^2 \\ {\left|z\right|}^4+\frac{1}{{\left|z\right|}^4}+2=a^2+b^2 \end{gathered}$$

For option (a),

$$\begin{gathered} \left|z\right|={\left(\frac{43+3\sqrt{205}}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\Rightarrow {\left|z\right|}^4=\left(\frac{43+3\sqrt{205}}{2}\right) \\ \frac{1}{{\left|z\right|}^4}=\frac{2}{43+3\sqrt{205}}=\left(\frac{43-3\sqrt{205}}{2}\right) \\ {\left|z\right|}^4+\frac{1}{{\left|z\right|}^4}+2=45\Rightarrow a^2+b^2=45 \end{gathered}$$

Only possible  if  $a=\pm 3,b=\pm 6$ or  $a=\pm 6,b=\pm 3$
Option (a) is correct.
For option (b),
$$\begin{gathered} {\left|z\right|}^4=\frac{7+\sqrt{33}}{4}, \frac{1}{{\left|z\right|}^4}=\frac{4}{7+\sqrt{33}}=\frac{7-\sqrt{33}}{4} \\ {\left|z\right|}^4+\frac{1}{{\left|z\right|}^4}+2=\frac{7}{2}+2=\frac{11}{2}\Rightarrow a^2+b^2=\frac{11}{2} \end{gathered}$$

Sum of two integers cannot be  $\frac{11}{2}$,
So wrong option.
For option (c),
${\left|z\right|}^4=\frac{9+\sqrt{65}}{4},\frac{1}{{\left|z\right|}^4}=\frac{9-\sqrt{65}}{4}$

${\left|z\right|}^4+\frac{1}{{\left|z\right|}^4}+2=\frac{9}{2}+2=\frac{13}{2}\Rightarrow a^2+b^2=\frac{13}{2}$
Sum of two integers can not be  $\frac{13}{2}$. So wrong option.
For Option (d),
$$\begin{gathered} {\left|z\right|}^4=\frac{7+\sqrt{13}}{6},\frac{1}{{\left|z\right|}^4}=\frac{7-\sqrt{13}}{6} \\ {\left|z\right|}^4+\frac{1}{{\left|z\right|}^4}+2=\frac{7}{3}+2=\frac{13}{3}\Rightarrow a^2+b^2=\frac{13}{3} \end{gathered}$$

Sum of two integers cannot be $\frac{13}{3}$. So wrong Option