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

` HTML block. Let's write out the clean answer and full step-by-step solution carefully. Clean answer: `${\left(\frac{43+3\sqrt{205}}{2}\right)}^{\frac{1}{4}}$` ${\left(\frac{43+3\sqrt{205}}{2}\right)}^{\frac{1}{4}}$

The correct option is ${\left(\frac{43+3\sqrt{205}}{2}\right)}^{\frac{1}{4}}$.

Step-by-step Explanation:

Let $z=r{e}^{i\theta }$, where $r=\left|z\right|>0$ and $\theta$ is the argument of $z$.

The complex conjugate of $z$ is given by:

$\bar{z}=r{e}^{-i\theta }$

Now, let us find the terms ${\left(\bar{z}\right)}^2$ and $\frac{1}{z^2}$:

${\left(\bar{z}\right)}^2=r^2{e}^{-2i\theta }$

$\frac{1}{z^2}=\frac{1}{r^2{e}^{2i\theta }}=\frac{1}{r^2}{e}^{-2i\theta }$

Adding these two expressions gives:

$w={\left(\bar{z}\right)}^2+\frac{1}{z^2}=\left(r^2+\frac{1}{r^2}\right)e^{-2i\theta }$

Using Euler's formula $e^{-2i\theta }=\cos \left(2\theta \right)-i\sin \left(2\theta \right)$, we separate $w$ into real and imaginary parts:

$\text{Re}\left(w\right)=\left(r^2+\frac{1}{r^2}\right)\cos \left(2\theta \right)$

$\text{Im}\left(w\right)=-\left(r^2+\frac{1}{r^2}\right)\sin \left(2\theta \right)$

We are given that both $\text{Re}\left(w\right)$ and $\text{Im}\left(w\right)$ are integers. Let:

$\text{Re}\left(w\right)=m\in \mathbb{Z}$

$\text{Im}\left(w\right)=n\in \mathbb{Z}$

Taking the sum of the squares of the real and imaginary parts:

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

Thus, we have:

${\left(r^2+\frac{1}{r^2}\right)}^2=m^2+n^2$

Expanding the left side:

$r^4+\frac{1}{r^4}+2=m^2+n^2$

$r^4+\frac{1}{r^4}=m^2+n^2-2$

For the option $r={\left(\frac{43+3\sqrt{205}}{2}\right)}^{\frac{1}{4}}$, we have:

$r^4=\frac{43+3\sqrt{205}}{2}$

To find $\frac{1}{r^4}$, rationalize the denominator:

$\frac{1}{r^4}=\frac{2}{43+3\sqrt{205}}=\frac{2(43-3\sqrt{205})}{{43}^2-9\times 205}=\frac{2(43-3\sqrt{205})}{1849-1845}=\frac{43-3\sqrt{205}}{2}$

Adding $r^4$ and $\frac{1}{r^4}$:

$r^4+\frac{1}{r^4}=\frac{43+3\sqrt{205}}{2}+\frac{43-3\sqrt{205}}{2}=43$

Substituting this back into the equation for $m^2+n^2$:

$m^2+n^2-2=43\Rightarrow m^2+n^2=45$

Since $45=6^2+3^2$, integer solutions exist for $m$ and $n$ (for example, $m=6$ and $n=-3$).

Therefore, $\left|z\right|={\left(\frac{43+3\sqrt{205}}{2}\right)}^{\frac{1}{4}}$ is a possible value of $\left|z\right|$.