Truly complex

Let $z = a + bi, \bar{z} = a - bi$ with $a,b \in \mathbb{Z}$ . The above equation can be factored according to:

$z\bar{z}(z^2 + \bar{z}^{2}) = (a^2 + b^2)(a^2 - b^2 + 2abi + a^2 - b^2 - 2abi) = 2(a^2 + b^2 )(a^2 - b^2) = 2(a^4 - b^4) = 350$ ;

or $a^4 - b^4 = 175$ . Knowing that $175 = 5^27^1$ we can now write $175 = (a^2 + b^2)(a^2-b^2)$ and test according to the factors:

$a^2 + b^2 = 175, 35, 25$

$a^2 - b^2 = 1, 5, 7$

which are satisfied for $a = \pm4, b = \pm3$ . Thus the vertices of the rectangle in question include the roots $z = 4 \pm 3i, -4 \pm 3i \Rightarrow$ a rectangle of side lengths $8$ and $6$ with area of $\boxed{48}.$