Question

On the complex plane, the parallelogram formed by the points $0, z, \frac{1}{z}$ and $z+\frac{1}{z}$ has area $\frac{35}{37}$. If the real part of $z$ is positive, let $d$ be the smallest possible value of $\left|z+\frac{1}{z}\right|$. Compute $\left[d^2\right]$.
[Note: [k] denotes the greatest integer function less than or equal to $k$.]

Answer 1

Area $= OA \times OC \times \sin 2 \alpha$
$=\frac{1}{| z |} \times| z | \sin 2 \alpha=\frac{35}{37} $
$\Rightarrow \sin (2 \alpha)=\frac{35}{37} \Rightarrow \cos 2 a =\frac{12}{37} $
$AM =| z | \cos 2 \alpha$
$BM =| z | \sin 2 \alpha$

In $\triangle OBM$
$\therefore OB ^2= OM ^2+ BM ^2$
$ \left| z +\frac{1}{ z }\right|^2=\left(\frac{1}{| z |}+| z | \cos 2 \alpha\right)^2+(| z | \sin 2 \alpha)^2$

$d ^2=\frac{1}{| z |^2}+| z |^2 \cos ^2(2 \alpha)+2 \cos 2 \alpha+| z |^2 \sin ^2 2 \alpha $
$d ^2=\frac{1}{| z |^2}+| z |^2+2 \cos (2 \alpha) \geq 2+2 \cos 2 \alpha=2+2\left(\frac{12}{37}\right)=\frac{98}{37}$