jmc

Let $a,$ $b,$ $c,$ $d$ be the roots of the quartic. Let $A$ be the point corresponding to complex number $a,$ etc.

Let $O$ be the center of the rhombus. Then the complex number corresponding to $O$ is the average of $a,$ $b,$ $c,$ $d.$ By Vieta's formulas, $a+b+c+d=-\frac{8i}{2}=-4i,$ so their average is $\frac{-4i}{4}=-i.$ Hence, $O$ is located at $-i.$



Let $p=OA$ and $q=OB.$ Then we want to compute the area of the rhombus, which is $4\cdot \frac{1}{2}pq=2pq.$

We see that $p=|a+i|=|c+i|$ and $q=|b+i|=|d+i|.$

Since $a,$ $b,$ $c,$ $d$ are the roots of the quartic in the problem, we can write $$2z^4+8i{z}^3+(-9+9i)z^2+(-18-2i)z+(3-12i)=2(z-a)(z-b)(z-c)(z-d).$$Setting $z=-i,$ we get $$4-3i=2(-i-a)(-i-b)(-i-c)(-i-d).$$Taking the absolute value of both sides, we get $$5=2|(a+i)(b+i)(c+i)(d+i)|=2p^2{q}^2.$$Then $4p^2{q}^2=10,$ so $2pq=\boxed{\sqrt{10}}.$