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 $$(\bar z)^2+\frac{1}{z^2}$$ are integers, then which of the following is/are possible value(s) of $\left|z\right|$
(A) $\left(\frac{43+3\sqrt{205}}{2}\right)^{\frac{1}{4}}$
(B) $\left(\frac{7+\sqrt{33}}{4}\right)^{\frac{1}{4}}$
(C) $\left(\frac{9+\sqrt{65}}{4}\right)^{\frac{1}{4}}$
(D) $\left(\frac{7+\sqrt{13}}{6}\right)^{\frac{1}{4}}$
My Attempt
Taking $z=x+iy$ we get $(\bar z)^2+\frac{1}{z^2}=(x^2-y^2)(1+\frac {1}{(x^2+y^2)^2})+i(2xy)(1+\frac{1}{(x^2+y^2)^2})$.
The calculation is appearing very arduous. Can there be a simple approach
Best Answer
Let $r = |z|$ be a possible value of the modulus. We claim that the above condition is equivalent to $$ \frac{(|z|^4 +1)^2}{|z|^4} = a^2 + b^2; \ \ a,b \in \mathbb{Z} $$ It is clear that the condition is necesary, because this the left side is just $|f(z)|^2$. To prove the other implication, consider the map $f(z) = \bar{z}^2 + \frac{1}{z^2} = \frac{1 + |z|^4}{z^2}$. Let $z = |z|e^{\frac{iarg(a-ib)}{2}}$. Then, $f(z) = a + ib$. Then $|f(z)| = \sqrt{a^2 + b^2}$ and $arg(f(z)) = arg(a +ib)$, so $f(z) = a + ib$. Thus, the condition is sufficient.