Given area of parallelogram, find smallest distance from origin to furthest point (complex numbers) 
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 $d^2.$

I begin by letting $z=x+yi$ and $\frac{1}{z}=\frac{x-yi}{x^2+y^2}$. Then, I applied the cross product to get that the
$$\text{Area of Parallelogram} = |z| \left|\frac{1}{z}\right| \sin(\theta) . $$
This implies that
$$ \frac{35}{37}=\left(\sqrt{x^2+y^2}\right) \cdot \left(\frac{1}{\sqrt{x^2+y^2}}\right) \sin(\theta) \Longrightarrow \frac{35}{37} = \sin(\theta) , $$
where $\theta$ is the angle between the $\frac{1}{z}$ and $z$. Subsequently, $\sin\left(\frac{\theta}{2}\right)=\frac{5}{\sqrt{74}}$ meaning that $z$ is in the form $x+\frac{5x}{7}i$. The problem reduced to an optimization problem, minimizing
$$ \left(x+\frac{x}{x^2+y^2}\right)^2+\left(y-\frac{y}{x^2+y^2}\right)^2=\left(x+\frac{x}{\frac{x\sqrt{74}}{7}}\right)^2+\left(\frac{5x}{7}-\frac{\frac{5x}{7}}{\frac{x\sqrt{74}}{7}}\right)^2 . $$
I eventually got the answer to be $\left(\frac{35}{37}\right)^2=\frac{1225}{1369}$. However, this answer is wrong. I am not sure where I messed up. Where did I go wrong?

Let $a$ and $b$ two $2$-dimensional vectors satisfying $||a||\cdot||b||=1$.  Then (via Gram) the square of area of the parallelogram spanned by these vectors is $(35/37)^2=||a||^2\cdot||b||^2-\langle a,b\rangle^2=1-\langle a,b\rangle^2$ from where $\langle a,b\rangle=\pm12/37$.
Now
$$\begin{align}
d^2=||a+b||^2&=||a||^2+||b||^2+2\langle a,b\rangle\\
&=||a||^2+\frac{1}{||a||^2}\pm\frac{12}{37}\\
&=\left(||a||+\frac{1}{||a||}\right)^2-2
\pm\frac{12}{37}\\
&\geq 4-2
\pm\frac{12}{37}\\
&=2
\pm\frac{12}{37},
\end{align}$$
as for all reals $x$ we know $-2<x+1/x<2$.
