Possible Modulus Values of Non-Zero Complex Number with Integer Parts

Let $z = r e^{i\theta}$ be a non-zero complex number, where $r = |z| > 0$ and $\theta \in \mathbb{R}$.

The complex conjugate of $z$ is $\bar{z} = r e^{-i\theta}$. We can evaluate the given expression: $(\bar{z})^2 + \frac{1}{z^2} = (r e^{-i\theta})^2 + \frac{1}{(r e^{i\theta})^2} = r^2 e^{-2i\theta} + \frac{1}{r^2} e^{-2i\theta} = \left( r^2 + \frac{1}{r^2} \right) e^{-2i\theta}$

Using Euler's formula, $e^{-2i\theta} = \cos(2\theta) - i \sin(2\theta)$, so: $(\bar{z})^2 + \frac{1}{z^2} = \left( r^2 + \frac{1}{r^2} \right) \cos(2\theta) - i \left( r^2 + \frac{1}{r^2} \right) \sin(2\theta)$

We are given that both the real and imaginary parts of $(\bar{z})^2 + \frac{1}{z^2}$ are integers. Let: $\text{Re}\left( (\bar{z})^2 + \frac{1}{z^2} \right) = m \in \mathbb{Z}$ $\text{Im}\left( (\bar{z})^2 + \frac{1}{z^2} \right) = n \in \mathbb{Z}$

Thus, $m = \left( r^2 + \frac{1}{r^2} \right) \cos(2\theta)$ $n = -\left( r^2 + \frac{1}{r^2} \right) \sin(2\theta)$

Squaring and adding these two equations: $m^2 + n^2 = \left( r^2 + \frac{1}{r^2} \right)^2 \left( \cos^2(2\theta) + \sin^2(2\theta) \right) = \left( r^2 + \frac{1}{r^2} \right)^2$

Let $N = m^2 + n^2$. Since $m, n \in \mathbb{Z}$, $N$ must be a non-negative integer that can be represented as the sum of two squares of integers.

Taking the square root on both sides: $r^2 + \frac{1}{r^2} = \sqrt{N}$

We can express this as a quadratic equation in $r^2$: $r^4 - \sqrt{N} r^2 + 1 = 0$

Solving for $r^2$: $r^2 = \frac{\sqrt{N} \pm \sqrt{N - 4}}{2}$

Squaring both sides to find $r^4$: $r^4 = \left( \frac{\sqrt{N} \pm \sqrt{N - 4}}{2} \right)^2 = \frac{N + (N - 4) \pm 2\sqrt{N(N - 4)}}{4} = \frac{N - 2 \pm \sqrt{N^2 - 4N}}{2}$

Thus, the possible values of $|z| = r$ are given by: $|z| = \left( \frac{N - 2 \pm \sqrt{N^2 - 4N}}{2} \right)^{\frac{1}{4}}$ where $N = m^2 + n^2 \in \mathbb{Z}$.

Now, let $K = |z|^4 = r^4$. Note that: $N = \left( r^2 + \frac{1}{r^2} \right)^2 = r^4 + 2 + \frac{1}{r^4} = K + 2 + \frac{1}{K}$ For $z$ to exist, $N$ MUST be an integer. Let's test each option for $K = |z|^4$:

1. Option (A): $|z| = \left(\frac{43+3\sqrt{205}}{2}\right)^{\frac{1}{4}}$ Here, $K = \frac{43 + \sqrt{1845}}{2}$. $\frac{1}{K} = \frac{2}{43 + \sqrt{1845}} = \frac{2(43 - \sqrt{1845})}{1849 - 1845} = \frac{43 - \sqrt{1845}}{2}$ $N = K + 2 + \frac{1}{K} = \frac{43 + \sqrt{1845}}{2} + 2 + \frac{43 - \sqrt{1845}}{2} = 43 + 2 = 45$ Since $N = 45 = 6^2 + 3^2$, $N$ is indeed an integer formed by the sum of two squares ($m=6, n=3$). Therefore, Option (A) is a possible value of $|z|$.

2. Option (B): $|z| = \left(\frac{7+\sqrt{33}}{4}\right)^{\frac{1}{4}}$ Here, $K = \frac{7 + \sqrt{33}}{4}$. $\frac{1}{K} = \frac{4}{7 + \sqrt{33}} = \frac{4(7 - \sqrt{33})}{49 - 33} = \frac{7 - \sqrt{33}}{4}$ $N = K + 2 + \frac{1}{K} = \frac{7 + \sqrt{33}}{4} + 2 + \frac{7 - \sqrt{33}}{4} = \frac{14}{4} + 2 = \frac{11}{2} \notin \mathbb{Z}$ Since $N$ is not an integer, Option (B) is not possible.

3. Option (C): $|z| = \left(\frac{9+\sqrt{65}}{4}\right)^{\frac{1}{4}}$ Here, $K = \frac{9 + \sqrt{65}}{4}$. $\frac{1}{K} = \frac{4}{9 + \sqrt{65}} = \frac{4(9 - \sqrt{65})}{81 - 65} = \frac{9 - \sqrt{65}}{4}$ $N = K + 2 + \frac{1}{K} = \frac{9 + \sqrt{65}}{4} + 2 + \frac{9 - \sqrt{65}}{4} = \frac{18}{4} + 2 = \frac{13}{2} \notin \mathbb{Z}$ Since $N$ is not an integer, Option (C) is not possible.

4. Option (D): $|z| = \left(\frac{7+\sqrt{13}}{6}\right)^{\frac{1}{4}}$ Here, $K = \frac{7 + \sqrt{13}}{6}$. $\frac{1}{K} = \frac{6}{7 + \sqrt{13}} = \frac{6(7 - \sqrt{13})}{49 - 13} = \frac{7 - \sqrt{13}}{6}$ $N = K + 2 + \frac{1}{K} = \frac{7 + \sqrt{13}}{6} + 2 + \frac{7 - \sqrt{13}}{6} = \frac{14}{6} + 2 = \frac{13}{3} \notin \mathbb{Z}$ Since $N$ is not an integer, Option (D) is not possible.

Hence, the only possible value for $|z|$ among the given options is (A).