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

$p(x)=0:\ \text{real coefficients, purely imaginary roots}$

$\text{so the roots are } \pm i\beta\ \ (\beta \in \mathbb{R})$

$\text{To find: the nature of the roots of } p(p(x))=0$

$1.\ \text{general form of } p \text{ from its purely imaginary roots}$

$2.\ \text{solve } p(p(x))=0 \text{ for } p(x)$

$3.\ \text{decide the nature of each root } x$

$\text{roots } \pm i\beta \Rightarrow p(x)=a(x-i\beta)(x+i\beta)$

$p(x)=a(x^2+\beta^2),\quad a\in\mathbb{R}\setminus\{0\}$

$p(p(x))=a\big((p(x))^2+\beta^2\big)$

$p(p(x))=0\ \Leftrightarrow\ (p(x))^2=-\beta^2$

$\Leftrightarrow\ p(x)=\pm i\beta$

$a(x^2+\beta^2)=i\beta$

$x^2=-\beta^2+i\dfrac{\beta}{a}$

$a(x^2+\beta^2)=-i\beta$

$x^2=-\beta^2-i\dfrac{\beta}{a}$

$x^2=w,\quad \operatorname{Im}w=\pm\dfrac{\beta}{a}\neq 0$

$x\in\mathbb{R}\Rightarrow x^2\ge 0;\qquad x=iy\Rightarrow x^2=-y^2\le 0$

$\text{so } x \text{ is neither real nor purely imaginary}$

$\text{Check: } p(x)=x^2+1\ (\beta=1,\ a=1)$

$p(p(x))=(x^2+1)^2+1=x^4+2x^2+2$

$x^2=-1\pm i\ \Rightarrow\ \text{roots non-real, non-purely-imaginary}$

$\text{Answer: (4) neither real nor purely imaginary roots}$