Roots of Composite Quadratic Function with Purely Imaginary Roots

To find the nature of the roots of the equation $p(p(x)) = 0$, let us analyze the given quadratic polynomial $p(x)$.

Let $p(x) = ax^2 + bx + c$, where $a, b, c \in \mathbb{R}$ and $a \neq 0$.

It is given that the quadratic equation $p(x) = 0$ has purely imaginary roots. Since the coefficients are real, the roots must occur in complex conjugate pairs. Thus, the roots of $p(x) = 0$ can be written as: $\alpha = ik \quad \text{and} \quad \beta = -ik$ where $k \in \mathbb{R}$ and $k \neq 0$.

Therefore, the polynomial $p(x)$ can be expressed as: $p(x) = a(x - ik)(x + ik) = a(x^2 + k^2)$

Now, consider the composite equation: $p(p(x)) = 0$

For $p(p(x)) = 0$, the value of $p(x)$ must be equal to the roots of $p(x) = 0$. Hence: $p(x) = ik \quad \text{or} \quad p(x) = -ik$

Substituting the expression for $p(x)$: $a(x^2 + k^2) = \pm ik$

Dividing by $a$ (since $a \neq 0$): $x^2 + k^2 = \pm \frac{ik}{a}$ $x^2 = -k^2 \pm i\left(\frac{k}{a}\right)$

Since $k \neq 0$ and $a \in \mathbb{R}$, the imaginary part of $x^2$ is non-zero: $\text{Im}(x^2) = \pm \frac{k}{a} \neq 0$

Let $x = u + iv$, where $u, v \in \mathbb{R}$. Then: $x^2 = (u + iv)^2 = (u^2 - v^2) + 2iuv$

Comparing the imaginary parts: $2uv = \pm \frac{k}{a} \neq 0$

This immediately implies that:

1. $u \neq 0$, which means that $x$ cannot be purely imaginary.
2. $v \neq 0$, which means that $x$ cannot be purely real.

Thus, all four roots of the equation $p(p(x)) = 0$ have both non-zero real parts and non-zero imaginary parts.

Therefore, the equation has neither real nor purely imaginary roots.

Correct Answer: (D)