$$|z^2||z-\bar z||z+\bar z|=1200$$ $$(x^2+y^2)(|2x|)(|2y|)=1200$$
$$(x^2+y^2)(|xy|)=300$$
I don’t think there is any realistic way to obtain $x$ and $y$ other than hit and trial. I am asking this question to know if there is one. Thanks!
If not possible, then any integers which satisfy this can be provided as an answer.
Answer is 62
Please note: $x,y$ are integers.
We have $$|z^2-\bar z^2| = |x^2 + 2ixy - y^2 -(x^2 - 2ixy - y^2)| = |4ixy| = 4xy$$ And $$z \bar z = (x+iy)(x-iy) = x^2+y^2$$So $$(x^2+y^2)|4xy| = 1200 \implies (x^2+y^2)|xy| = 300$$ Note that if $(x,y)$ is solution then $(x , -y) , (-x ,y) , (-x , -y), (y , x) , (-y , x) , (y ,-x)$ and $(-y , -x)$ are also solution as well. Because $x , y \in \mathbb{Z}$ we have $x^2 + y^2 \in \mathbb{N}$ and $|xy| \in \mathbb{N}$. The positive divisors of $300$ is as follows: $$1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300$$
We see that $12 \times 25 = 300$ and also $x =3 , y = 4$ works, so we have all $8$ points because the other points are $(4,3) , (-3 , -4) , (-4 , -3) , (-3 , 4) , (3 , -4) , (4,-3) , (-4 , 3)$.
You can compute the area by adding the areas of rectangle and two trapezoids $6\times 8 + 2\times(\frac{(6 + 8)1}{2}) = 48 + 14 = 62$.
