JEE-Advanced 2014 (2/40)

Since $p(x)$ has purely imaginary roots, we can write it as: $p(x)=(x+ai)(x-ai)=x^2+a^2$

Here, $a\in R$ and $a\ne 0$

Now, $p(p(x))=x^4+2(ax)^2+a^4+a^2$

This is a bi-quadratic equation and can be solved by quadratic formula: ${ x }^{ 2 }=\frac { -2{ a }^{ 2 }\pm \sqrt { 4{ a }^{ 4 }-4{ a }^{ 4 }-4{ a }^{ 2 } } }{ 2 } =-{ a }^{ 2 }\pm \sqrt { -{ a }^{ 2 } }$

Since $a\in R$ , $a^2>0$ . Hence the above equation tells that $p(p(x))$ has neither real nor purely imaginary roots.

Since, the p(x) has purely imaginary roots, hence sum of the roots will be zero (It is given in question that coefficients are real). Hence , middle term of p(x) is zero. equation can be considered as ax^2+c. So , for any real and purely imaginary value of x, p(x) will return a real value. Given that p(x) has real roots, p(p(x)) will surely have neither real nor purely imaginary roots.

Focus of the word purely imaginary, p(x) and p(p(x)) both may have complex roots but not purely imaginary.