Proof of complex conjugates of a polynomial

I was trying to prove the following:

"In an equation with real coefficients, the non-real roots occur in conjugate pairs"

While proving this, I reached a contradiction and I don't understand what mistake I made.

My proof is as follows:

let $f(x) = 0$ be an equation with real roots. let $x = a+bi$ be a root of $f(x)$ such that $f(a+bi) = 0$

Now from the factor theorem, we can say that

$$f(x) = (x - (a+bi)) \cdot Q(x)$$

where $Q(x)$ is the quotient of the division

Rearranging this equation,

$$f(x) = (x - a - bi) \cdot Q(x)$$

$$f(x) = (x-a) \cdot Q(x) - bi \cdot Q(x)$$

if we divided $f(x)$ with $(x - (a - bi))$, from the remainder theorem, we get

$$f(x) = (x - (a - bi)) \cdot \phi(x) + R(x)$$

where $\phi(x)$ is the quotient and R(x) is the remainder

Rearranging this as well

$$f(x) = (x - a + bi) \cdot \phi(x) + R(x)$$

$$f(x) = (x - a) \cdot \phi(x) + bi \cdot \phi(x) + R(x)$$

$$f(x) = (x - a) \cdot \phi(x) + R(x) + bi\cdot \phi(x)$$

thus, we can say

$$f(x) = f(x)$$

$$(x-a) \cdot Q(x) - bi \cdot Q(x) = (x - a) \cdot \phi(x) + R(x) + bi \cdot \phi(x)$$

equating the imaginary parts, we get

$$-b \cdot Q(x) = b \cdot \phi(x)$$

$-Q(x) = \phi(x) \rightarrow$ (eq1)

and the real parts

$$(x-a) \cdot Q(x) = (x-a) \cdot \phi(x) + R(x)$$

$$(x-a) \cdot (Q(x)-\phi(x)) = R(x)$$

Now, this goes out of the proof, but if we assume that a-bi is also a root of f(x) (which it is), R(x) being the remainder, must be 0 as f(x) is divisible by (x-(a-bi))

thus we get,

$$(x-a) \cdot(Q(x)-\phi(x)) = 0$$

Either $(x-a) = 0$ or $Q(x) - \phi(x) = 0$

For all $x \neq a$, we get $Q(x) - \phi(x) = 0$ or $$Q(x) = \phi(x)$$

Which contradicts with eq 1

Thank you!

Solution 1:

Even if $x$ is a real number, you cannot assume that $Q(x)$ is such a number. The coefficients of $f(x)$ are real, but $Q(x)=\frac{f(x)}{x-a-bi}$, and so, in general, its coefficients will not be real. So, think in terms of “real part” and “imaginary part” leads to errors.

For instance, if $a=1$, $b=1$, $c=2i$, and $d=-1-i$, then $a+bi=c+di$. However, $a\ne c$ and $b\ne d$.