Complex Rhombus

let $f(z) = 2z^4 + 8i z^3 + (-11+7i) z^2 + (-14-6i) z + 25-14i = 2(z-z_1)(z-z_2)(z-z_3)(z-z_4)$ where $z_k$ are the roots.

sum of roots $= \frac{-8i}{2} = -4i$ , average of roots $= \frac{-4i}{4} = -i$ which gives the center of the rhombus.

$a^2 b^2 = |(-i-z_1)(-i-z_2)(-i-z_3)(-i-z_4)| = |\frac{1}{2}f(-i)| = |12 - \frac{7}{2}i| = \frac{25}{2} \implies ab = \sqrt{\frac{25}{2}}$

area of rhombus $= 2ab = \sqrt{50} \implies \therefore A=50$