Will the roots of $p(p(x))=0$ be purely real or purely imaginary or neither?

Question

Question

The quadratic equation $p(x)=0$ with real coefficients has purely imaginary roots. Then the equation $p(p(x))=0$ will have -

a) only purely imaginary roots

b) all real roots

c) two real and two imaginary roots

d) neither real nor purely imaginary roots

My Thoughts

I assumed a quadratic equation $${p(x)=ax^2+bx+c=0}$$ Now, as the coefficients are real the two roots must be conjugates. Let these roots be $ki$ and $-ki$. Satisfying it in the equation gives $b=0 \Rightarrow p(x)=ax^2+c=0$ or $x^2+λ=0$ where $λ=c/a$ (to reduce the variables). Now, putting it in $p(p(x))=0$ we get $x^4+2λx^2+λ^2+λ=0$. Now I am stuck here. I don't know how to proceed further.

Answer 1

WLOG the polynomial has the form $p(x)=x^2+a^2$. Then

$$(x^2+a^2)^2+a^2=0$$ has two distinct pairs of complex conjugate roots.