Do complex roots always come in pairs? The Complex Conjugate Theorem states that, given the polynomial $p(x)$ with real coefficients $p \in \mathbb{R}[p]$ and one of its roots being $a+bi \in \mathbb{C}$, its complex-conjugate pair $\overline {z}$ must be a root as well. We could also intuitively conclude that only by multiplying $(a+bi)$ with $(a-bi)$ will the imaginary parts be destroyed, allowing $p$ to be a real polynomial.  
Following this line of reasoning, it shouldn't be necessary for complex roots of a complex polynomial to come in pairs. However, this premise was used in solving one of my school assignments. 
Do complex roots always have to come in pairs, regardless of the field in which the polynomial was defined?
 A: $(z-i)(z-2i)$ has two complex roots, $i$ and $2i$, but their conjugates are not roots. 
A: Consider the polynomial $z-i$ . . .

It's good to remember the reason the complex conjugate theorem is true in the first place: the map $z\mapsto \overline{z}$ is an automorphism of the field $\mathbb{C}$, and fixes the subfield $\mathbb{R}$ (this is a fancy way of saying $\overline{r}=r$ if $r$ is real). Thus - defining the "conjugate" $\overline{p}$ of a polynomial (with coefficients in $\mathbb{C}$) to be the polynomial whose coefficients are the conjugates of the corresponding coefficients of $p$ - we have the following:


*

*If $p$ is any polynomial and $z$ is a root of $p$, then $\overline{z}$ is a root of $\overline{p}$.

*If the coefficients of $p$ are from $\mathbb{R}$, then $\overline{p}=p$.
This second fact is no longer true for polynomials with coefficients from outside $\mathbb{R}$!
Note that this generalizes in a natural way to:

Suppose $F$ is any subfield of $\mathbb{C}$, $\varphi$ is a field automorphism of $\mathbb{C}$ which fixes $F$, and $p$ is a polynomial with coefficients from $F$. Then if $z$ is a root of $p$, so is $\varphi(z)$.

This is the (fine, "a") first step towards Galois theory . . .
