Finding $a$ such that the complex solutions of $z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0$ form a parallelogram in the complex plane Find all values of the real number $a$ so that the four complex roots of
$$z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0$$ form the vertices of a parallelogram in the complex plane.

I set $z = x + yi$ due to the given knowledge that $z$ is complex. I then substituted it in, which gave me $$(x+yi)^4 - 6(x+yi)^3 + 11a(x+yi)^2 - 3(2a^2 + 3a - 3)(x+yi) + 1 = 0.$$ However, I'm afraid it's a bit too bashy for it to be viable, so are there any other ways I can get started on this question?

Expanding on @AlexeyBurdin's answer (in case anyone wants to see how to do it by hand), since $x=3/2$ we have $y^2+z^2=27/2-11a$ and$$6a^2+9a-9=27/2-3(y^2+z^2)=33a-27\implies a\in\{1,\,3\}.$$Each such $a$ works viz.$$1=x^4-(y^2+z^2)x^2+y^2z^2=81/16-(27/2-11a)9/4+y^2z^2,$$which we can simplify to obtain $y^2z^2$. We then have the sum and product of $y^2,\,z^2$, and solutions in $\Bbb C$ exist.
