How to prove that conjugate roots always come in pairs [duplicate]

As noted, this is only true if the coefficients are real numbers. To prove it, it's enough to prove two lemmas: $\bar {z_1} + \bar {z_2} = \overline {z_1+z_2}$, and $\bar {z_1} \cdot \bar {z_2} = \overline {z_1 z_2}$. You can prove these facts by expressing $z_n$ in the form $x_n + i y_n$ and then doing the arithmetic by brute force.

Once you know these facts are true, you can prove (by induction on the degree of the polynomial, if necessary) that for any $z \in \Bbb C$ and any $p(x) \in \Bbb R[x]$ we have $p(\bar z) = \overline {p(z)}$, so $p(z) = 0 = \bar 0 = \overline {p(z)}=p(\bar z)$, which means that whenever $z$ is a root of $p(z)$, so too is $\bar z$.

To test your understanding, do you see why the proof fails if $p$ has one or more non-real coefficients?

Here's a solution that only uses some simple algebra:

Imagine we have some polynomial with real coefficients (the only way this works, as you mentioned):

$P(x) = \alpha_0 + \alpha_1x + \alpha_2x^2$

Here, $\alpha_0, \alpha_1,$ and $\alpha_2$ are all real numbers. Any quadratic equation has two roots, and we already know one is the complex root $a + bi$, with $b \neq 0$. Let's call the other root $c$, and try to find out what it is.

By factoring, we have:

$P(x) = (x-(a+bi))(x-c) = (x-a-bi)(x-c)$

Expanding,

$P(x) = x^2 - ax - bix -cx + ac + bci$

And grouping like terms:

$P(x) = x^2 - (a+bi+c)x + c(a+bi)$

Now it's more managable. First of all, we know $a + bi + c = \alpha_1$, which we defined to be a real number. This gives us some new info about $c$. We now know $c$ must at least include some term $-bi$ in order to cancel out the imaginary part $+bi$.

We still don't know the real part of $c$, so we can let $c = d - bi$, and try to find $d$. Using the last term of our equation: because $\alpha_0$ is real, $c(a+bi)$ must be real. We can plug in our new definition of $c$ and expand:

$\alpha_0 = c(a+bi) = (d-bi)(a+bi) = ad - abi +bdi-bi^2 = ad - abi + bdi + b^2$

Like before, we know that $\alpha_0$ is real, so the imaginary parts must cancel out. Using our variables, this means:

$-abi + bdi = 0 \Rightarrow bdi = adi$

Dividing by $bi$ on both sides (note: you can't divide by 0, hence why we set $b \neq 0$ at the start) yields $d = a$. We can finally plug this back in to $c$, the other root of our polynomial $P(x)$, to get:

$c = a - bi$

which is the conjugate of $a + bi$.