Multiple roots of one polynomial of degree 4

Question

I have the following specific polynomial of degree 4: \begin{align} P(x) &= (x_1^2 + x_2^2)^2 + a(x_1 + i x_2)^2(x_1^2 + x_2^2) + \overline{a} (x_1-ix_2)^2(x_1^2 + x_2^2)\\ &+b(x_1+ix_2)^4 + \overline{b}(x_1-ix_2)^4, \, x=(x_1,x_2)\in \mathbb{R}^2. \end{align} In addition, I know that coefficients $a,b$ are such that \begin{equation} P(x) \geq c(x_1^2+x^2_2)^2 , \, c > 0. \end{equation}

I have a conjecture that $P$ must have only simple (complex) roots. For example, if one will look for the similar polynomial of degree 2, i.e.: \begin{align} &Q(x) = (x_1^2 + x_2^2) + a(x_1 + i x_2)^2 + \overline{a} (x_1-ix_2)^2,\\ &Q(x) \geq c(x_1^2 + x_2^2) \end{align} then one can directly show that the corresponding roots will be only simple. And condition of positivity is used to have that the determinant is not zero.

Answer 1

Sorry to all. Question is not at all well posed. By simple characterstics I meant that in the decomposition of the polynomial it won't appear any monomial of the form: \begin{equation} (x_1 + z x_2)^k,\, z \in \mathbb{C}, \, k\geq 2. \end{equation} The case when $a = 0, b = 0$ is the obvious counterexample, so the conjecture is absolutely not true.

Answer 2

HINT

$$P(x_1, x_2) = \Re F(x) = |x|^4\Re G(y),$$ where $$F(x) = |x|^4 + 2a|x|^2x^2 + 2bx^4,\quad G(y) = 1 + ay + by^2,\quad y=\left(\dfrac x{|x|}\right)^2.$$ This means that every root $|y| \ge 1$ defines the infinity family of solutions $$x = r\sqrt y,\quad r \in\mathbb R.$$