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

Answer: 0002

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

In $△OBM$
$\therefore O{B}^2=O{M}^2+B{M}^2$
${\left|z+\frac{1}{z}\right|}^2={\left(\frac{1}{|z|}+|z|\cos 2\alpha \right)}^2+(|z|\sin 2\alpha {)}^2$

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