Mathematics - Complex Number Question with Solution | TestHub

Let $z=r.e^{i\theta }$

So, ${\left(\overline{z}\right)}^2+\frac{1}{z^2}=\left(r^2+\frac{1}{r^2}\right)e^{-2i\theta }=a+ib$ (say), given $a,b\in Z$

Now taking mod both side and squaring we get,

${\left(r^2+\frac{1}{r^2}\right)}^2=a^2+b^2$

$\Rightarrow r^8-\left(a^2+b^2-2\right)r^4+1=0$

$\Rightarrow r^4=\frac{\left(a^2+b^2-2\right)\pm \sqrt{{\left(a^2+b^2-2\right)}^2-4}}{2}$

Now comparing from option we get, for $a^2+b^2=45$ (i.e $\left(a,b\right)=\left(\pm 6,\pm 3\right)$ or $\left(\pm 3,\pm 6\right)$

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