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?

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
$$a=1,\quad a=3.$$
