Animated Solution for Mathematics - Quadratic Equations: The quadratic equation $p(x)=0$ with real coefficients has purely imaginary roots. Then the equation $p(p(x))=0$ has

In the complex plane, a quadratic polynomial $p(x)=A{x}^2+Bx+C$ with purely imaginary roots implies that the roots lie on the vertical axis. These roots must be of the form $i\alpha$ and $-i\alpha$ for some real $\alpha$.

By Vieta's formulas, the sum of the roots is $-\frac{B}{A}$. Since $i\alpha +(-i\alpha )=0$, it follows that $B=0$.

This simplifies our polynomial to the form $p(x)=A{x}^2+C$. Without loss of generality, we can normalize this to:

where $a$ is a positive constant. This function serves as our primary transformation.

We are tasked with finding the roots of the composite function $p(p(x))=0$. Substituting the expression for $p(x)$ into itself, we obtain:

To solve for $x$, we treat this as a quadratic equation in terms of $p(x)$. Isolating the squared term yields:

Taking the square root of both sides, and noting that $a>0$, we find:

We now isolate $x^2$ to examine the nature of the roots:

Observe that $x^2$ is a complex number with a non-zero real part ($-a$) and a non-zero imaginary part ($\pm \sqrt{a}$).

If $x$ were a real number, $x^2$ would be real. If $x$ were purely imaginary, $x^2$ would be a real, non-positive number.

Since $x^2$ possesses a non-zero imaginary component, $x$ itself must be a complex number with both real and imaginary parts. Therefore, the roots of $p(p(x))=0$ are neither purely real nor purely imaginary, but reside in the four quadrants of the complex plane.