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.
Let $x$ be the parallelogram center. Then the roots are $x+y,\,x-y,\,x+z,\,x-z$ for some $y,z$. Then, by Vieta's formulas: $$\begin{cases} 6=4x\\ 6x^2-y^2-z^2=11a\\ 4 x^3 - 2 x y^2 - 2 x z^2 = 3(2a^2+3a−3)\\ (x^2-y^2)(x^2-z^2)=1 \end{cases}$$ Although it is doable by hand, too lazy to perform it here, feed it to WA to get $$a=1,\quad a=3.$$
