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Question

Let $(\bar{z})^2+\frac{1}{z^2}=m+$ in, $m, n \in Z$

$(\bar{z})^2+\frac{\bar{z}^2}{|z|^2}= m + in$

$\Rightarrow\left( x ^2- y ^2\right)\left(1+\f

Vedclass · Accepted Answer

$\left(\frac{43+3 \sqrt{205}}{2}\right)^{\frac{1}{4}}$

Vedclass · Answer

Let $(\bar{z})^2+\frac{1}{z^2}=m+$ in, $m, n \in Z$

$(\bar{z})^2+\frac{\bar{z}^2}{|z|^2}= m + in$

$\Rightarrow\left( x ^2- y ^2\right)\left(1+\frac{1}{|z|^4}\right)= m$    $. . . . (1)$

$\&-2 xy \left(1+\frac{1}{|z|^2}\right)= n$    $. . . . (2)$

Equation $(1)^2+(2)^2$

$\left(1+\frac{1}{|z|^2}\right)^2\left[\left(x^2+y^2\right)^2\right]=m^2+n^2$

$\left(1+\frac{1}{|z|^2}\right)^2(|z|)^2=m^2+n^2$

$\Rightarrow|z|^2+\frac{1}{|z|^2}+2=m^2+ n ^2$

Now for option $(A)$

$|z|^4=\frac{43+3 \sqrt{205}}{2}$

$\Rightarrow m^2+ n ^2=45$

$\Rightarrow m = \pm 6, n = \pm 3$

Option $(B)$

$|z|^2+\frac{1}{|z|^2}+2=\frac{7+\sqrt{33}}{4}+\frac{7-\sqrt{33}}{4}+2=\frac{7}{2}+2=\frac{11}{2}$

Option $(C)$

$|z|^2+\frac{1}{|z|^2}+2=\frac{9+\sqrt{65}}{4}+\frac{9-\sqrt{65}}{4}+2=\frac{18}{4}+2=\frac{9}{2}+2=\frac{13}{2}$

Option $(D)$

$|z|^4+\frac{1}{|z|^2}+2=\frac{7+\sqrt{13}}{6}+\frac{7-\sqrt{13}}{6}+2=\frac{14}{6}+2=\frac{7}{3}+2=\frac{13}{2}$

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 $|z|$ ?

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