Given, a quadratic equation $$p(x)=0$$ with real coefficients has purely imaginary roots.

Let $$i \lambda$$ and $$-i \lambda$$ are the roots of $$p(x)=0$$ where $$i=\sqrt{-1}$$ and $$\lambda$$ is a real number except zero.

$$\begin{aligned} & \therefore p(x)=a(x-i \lambda)(x+i \lambda) \\ & \Rightarrow p(x)=a\left(x^2+\lambda^2\right) \end{aligned}$$

Now, $$\quad p(p(x))=0$$

$$\Rightarrow \quad a\left((p(x))^2+\lambda^2\right)=0$$

$$\Rightarrow a\left[a^2\left(x^2+\lambda^2\right)^2+\lambda^2\right]=0$$

$$\Rightarrow a^2\left(x^2+\lambda^2\right)^2+\lambda^2=0$$

$$\Rightarrow \quad\left(x^2+\lambda^2\right)^2=-\frac{\lambda^2}{a^2}$$

$$\Rightarrow \quad x^2+\lambda^2= \pm i \frac{\lambda}{a}$$

$$\Rightarrow \quad x^2= \pm i \frac{\lambda}{a}-\lambda^2$$

$$\Rightarrow \quad x= \pm \sqrt{ \pm i \frac{\lambda}{a}-\lambda^2}$$

Hence, $$p(p(x))=0$$ has four roots but all the roots are neither purely real nor purely imaginary.

Hint:

(i) In a quadratic equation imaginary roots are always in conjugate pair i.e, if one root is $$p+i q$$, then other root must be $$p-i q$$

(ii) A quadratic equation with real coefficients has purely imaginary roots, then consider $$\pm i \lambda$$ are the roots of the given quadratic, where $$\lambda \in \mathrm{R}-\{0\}$$.