# Can roots of an algebraic equation be neither real nor complex?

My usually understanding tells me that an algebraic equation can intersect the x-axis, touch it, or not touch it. In which cases I will get a real root, coinciding real root, and an imaginary root (conjugate pair) respectively.
Note that I am considering real coefficients only.
But consider the following:

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

Since $p(x)$ can become zero only on input of some complex numbers, I get the sense that the roots of the equation $p(p(x)) = 0$ will also be imaginary. But this is all based on intuition.

The answer provided to me was that The roots are neither real nor purely imaginary. Then what are the roots? I mean they must be either real or imaginary.

All help will be appreciated!

• You said "the roots are neither real nor purely imaginary" . That means roots with both a real part and an imaginary part are present. – The Integrator Apr 29 '18 at 18:24
• A complex number has the form $a+bi$ where $a,b$ are real number variables. Ordinarily, if $b=0$ then we abbreviate $a+0\cdot i = a$ and that's a real number. Also, if $a=0$ then we abbreviate $0+bi=bi$ and that's a purely imaginary number. For example, the roots of $p(x)=x^2+x+1$ are $-\frac{1}{2} + \frac{\sqrt{3}}{2} i$ and $-\frac{1}{2} - \frac{\sqrt{3}}{2} i$, neither of which is real, and neither of which is purely imaginary. – Lee Mosher Apr 29 '18 at 18:27

This is just a vocabulary convention...

While a real number is $r$ in $(-\infty,\infty)$, a purely imaginary number is a product $r\cdot i$.

If we are using the notation $a+bi$ for arbitrary complex numbers, the purely real numbers have $b=0$ while the purely imaginary numbers have $a=0$.

For $p(x)=x^2+4$, the roots of $p(x)$ are purely imaginary: $\{-2i,+2i\}$. Not all of the roots of the equation $p(p(x))$ are purely imaginary.

• +1 I see. The answer provided to me was just worded badly. – SmarthBansal Apr 29 '18 at 18:34

To elaborate, if $$p(x)$$ is quadratic, then it must have two roots by The Fundamental Theorem of Algebra.

So we know $$p(x)=A(x-mi)(x-ni)$$ for some real numbers A,m,and n.

We know something more, m=-n by the Complex Conjugate Root Theorem, so $$p(x)=A(x-mi)(x+mi)$$.

What do you get if you multiply that out?

Now plug that result into itself and you get a quartic equation.

Do you know how to find these roots? The quadratic formula might be applicable.