$p(x)=0$ with real coefficient has purely Imaginary roots.Then the equation $p(p(x)) = 0$ has

If a quadratic equation has purely imaginary roots, then it can be written in the form $p(x)=k(x-ai)(x+ai)=k(x^2+a^2)$ for some $k,a\ne0$.

Hence $p(p(x))=k[k^2(x^2+a^2)^2+a^2]=k^3x^4+2k^3a^2x^2+k^3a^4+ka^2$. By the quadratic equation, this gives us $$x^2=\frac{-2k^3a^2\pm\sqrt{4k^6a^4-4k^3(k^3a^4+ka^2)}}{2k^3}=\frac{-2k^3a^2\pm\sqrt{-4k^4a^2}}{2k^3}=\frac{-ka^2\pm ia}{k}$$

This means that $x$ can be neither purely real or purely imaginary, since if it were, then $x^2$ would be real.

If $p(r) = 0$ it must be that $p(p(x)) = 0$ when $p(x) = r$ or when $p(x)$ equals one of it's roots.

The only (quadratic) polynomial with purely imaginary roots is $p(x) = ax^2 + b$.

If we let the roots $r = \pm iq$ then the solution to $p(x) = ax^2 + b =\pm iq$ cannot be pure real or pure imaginary.

If $x$ were pure real or pure imaginary the left hand side would be real so there would be no way to solve the equation.

The general form for p(x) is:

$p(x)=k(x-ai)(x+ai)=\\=kx^{2}+ka^2$

where k and a are real numbers.