Let $$\bar{z}$$ 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 $$(\bar{z})^2 + \dfrac{1}{z^2}$$ are integers, then which of the following is/are possible value(s) of $$|z|$$?

Let $$z$$ be written in polar form as $$z = r\,e^{i\theta} \;(r \gt 0,\, -\pi \lt \theta \le \pi)$$. Then $$\bar z = r\,e^{-i\theta}$$, so

$$ (\bar z)^2 + \frac{1}{z^{2}} = r^{2}e^{-i2\theta} + \frac{1}{r^{2}}e^{-i2\theta} = \bigl(r^{2}+r^{-2}\bigr)\,e^{-i2\theta}. $$

Put $$s = r^{2}+r^{-2} \; (>2).$$ Writing $$e^{-i2\theta} = \cos 2\theta - i\sin 2\theta,$$ we get

$$ (\bar z)^2 + \frac{1}{z^{2}} = s\cos2\theta \;-\; i\,s\sin2\theta. $$

The real part $$s\cos2\theta$$ and the imaginary part $$-\,s\sin2\theta$$ are both required to be integers. Hence there exist integers $$A,B$$ such that

$$ s\cos2\theta = A,\qquad -s\sin2\theta = B. $$

Using $$\cos^{2}2\theta+\sin^{2}2\theta = 1$$, we obtain

$$ \frac{A^{2}+B^{2}}{s^{2}} = 1 \;\;\Longrightarrow\;\; s^{2}=A^{2}+B^{2}. \tag{-1} $$

Thus $$s=\sqrt{A^{2}+B^{2}}$$ must itself be the square root of a sum of two perfect squares.

Because $$s = r^{2}+r^{-2},$$ we have the quadratic in $$r^{2}$$:

$$ (r^{2})^{2} - s\,r^{2} + 1 = 0 \;\;\Longrightarrow\;\; r^{2} = \frac{s \pm \sqrt{s^{2}-4}}{2}. $$

Taking the positive root and then $$r=\sqrt{r^{2}}$$ gives the admissible moduli $$|z|$$ for every allowed $$s$$.

We now test the four given options.

Let $$r = \Bigl(\dfrac{43+3\sqrt{205}}{2}\Bigr)^{1/4}.$$ Then

$$ r^{2} = \sqrt{\dfrac{43+3\sqrt{205}}{2}} = y \quad(\text{say}). $$

Compute $$s = y+\dfrac{1}{y}$$:

$$\begin{aligned} y^{2} &= \dfrac{43+3\sqrt{205}}{2},\\[4pt] \frac{1}{y^{2}} &= \dfrac{2}{43+3\sqrt{205}},\\[4pt] y^{2}+\frac{1}{y^{2}} &= \dfrac{43+3\sqrt{205}}{2} + \dfrac{2}{43+3\sqrt{205}} = 43, \\[4pt] s &= y+\frac{1}{y} = \sqrt{\,y^{2}+\frac{1}{y^{2}}+2\,} = \sqrt{43+2} = \sqrt{45} = 3\sqrt5. \end{aligned}$$

Because $$s^{2} = 45 = 6^{2}+3^{2},$$ equation $$-(1)$$ is satisfied with $$A=6,\;B=3.$$ Both parts of $$(\bar z)^2 + 1/z^{2}$$ can therefore be integral. Hence Option A gives a valid modulus.

Take $$r = \Bigl(\dfrac{7+\sqrt{33}}{4}\Bigr)^{1/4}.$$ Here

$$ y^{2} = \dfrac{7+\sqrt{33}}{4},\qquad y^{2}+\dfrac{1}{y^{2}} = \dfrac72,\qquad s = \sqrt{\dfrac72 + 2} = \sqrt{\dfrac{11}{2}}. $$

Then $$s^{2} = \dfrac{11}{2}$$ is not an integer, so $$A,B$$ cannot be integers. Option B is not possible.

Take $$r = \Bigl(\dfrac{9+\sqrt{65}}{4}\Bigr)^{1/4}.$$ Similarly we get

$$ y^{2} = \dfrac{9+\sqrt{65}}{4},\qquad y^{2}+\dfrac{1}{y^{2}} = \dfrac92,\qquad s^{2} = \dfrac{13}{2}, $$

which is not an integer. Option C is not possible.

Take $$r = \Bigl(\dfrac{7+\sqrt{13}}{6}\Bigr)^{1/4}.$$ We find

$$ y^{2} = \dfrac{7+\sqrt{13}}{6},\qquad y^{2}+\dfrac{1}{y^{2}} = \dfrac73,\qquad s^{2} = \dfrac{13}{3}, $$

again non-integral, so integers $$A,B$$ do not exist. Option D is not possible.

Therefore the only modulus that fulfils the given condition is

$$|z| = \Bigl(\dfrac{43 + 3\sqrt{205}}{2}\Bigr)^{1/4}.$$

Option A which is: $$\left(\dfrac{43 + 3\sqrt{205}}{2}\right)^{1/4}$$