Extending to something more sinister

From the Complex conjugate root theorem we have: If a quadratic equation has real coefficients and complex roots, then the roots are always conjugate of each other.

Let $ai$ be a root of $p(x)=0$ and $-ai$ be the other root, where $a$ is a non-zero real number.

Therefore, $p(x)=(x-ai)(x+ai)$ .

$p(p(x))=0$ $(p(x)-ai)(p(x)+ai)=0$ $((x-ai)(x+ai)-ai)((x-ai)(x+ai)+ai)=0$ $(x^2+a^2-ai)(x^2+a^2+ai)=0$

Therefore, $x=\pm\sqrt{-a^2\pm ai}$ , and the four solutions are neither real nor purely imaginary.

From the Complex conjugate root theorem we have: If a quadratic equation has real coefficients and complex roots, then the roots are always conjugate of each other.

We are given that $p(x) = 0$ has purely imaginary roots, let one root be $ai$ where $a$ is a non-zero real number. Then the other root will be its conjugate: $-ai$ . We get $p(x) = (x - ai)(x + ai) = x^2 + a^2$ .

Now, $p(p(x)) = (x^2 + a^2)^2 + a^2$ . It will be zero when $p(x)$ will be $\pm ai$ .

$x^2 + a^2 = \pm ai$ : If $x$ were a real number or a purely imaginary number, then the LHS would be a purely real number, and the RHS would be a purely imaginary number, but this is a contradiction, so the roots are neither real nor purely imaginary.

By example: Let p $(x)=x^2+1$ , then p has roots $\pm i$ . The roots of $p(p(x))=(x^2+1)^2+1$ are $x=\pm \sqrt{-1\pm i }$ which are neither real nor purely imaginary.