If an polynomial has a complex root, is it necessary that its conjugate is also a root? If an polynomial has a complex root, is it necessary that the conjugate of that root is also a root of the polynomial? I am not sure about this. Please comment on this.
 A: Key Idea $ $ Conjugation $\rm\:w\,\mapsto\, \overline w\:$ preserves $\rm\:\color{#c00}{sums\,\ \&\,\ products}.\:$ and $\rm\:\color{#0a0}{fixes\ reals}\in\color{#0a0}{\Bbb R},\:$ therefore by induction, it preserves polynomials functions of such: $\rm\  \overline{f(w)} = f(\overline w),\ \ f(x)\in\color{#0a0}{\Bbb R}[x],\ $ having all $\,\rm\color{#0a0}{real}$ coefficients, since such polynomials are compositions of said basic operations. $ $ Explicitly
$\begin{eqnarray}
\rm \overline{f(w)}
&=&\rm \  \overline{a_n w^n +\,\cdots + a_1 w + a_0}\\
&=&\rm\, \overline{a_n w^n}\, +\,\cdots + \overline{a_1 w} + \overline a_0\ \ by\ \ \ \color{#c00}{\overline{x+y}\, =\, \overline x + \overline y}\ \ \ \forall\ x,y \in \Bbb C\\
&=&\rm\,  \overline a_n\,  \overline w^n+\,\cdots + \overline a_1\overline w + \overline a_0\ \ by\ \ \ \color{#c00}{\overline{x\, *\, y}\, =\, \overline x\, *\, \overline y}\ \ \forall\ x,y \in \Bbb C \\
&=&\rm\, a_n\, \overline w^n + \,\cdots + a_1 \overline w + a_0\ \ by\ \ \ \color{#0a0}{\overline a = a}\ \  \forall\ \color{#0a0}a\in \color{#0a0}{\Bbb R}\\
&=&\rm\  f(\overline w)\\
\rm\!\! So\ \ 0\! =\! f(w)\! \ \Rightarrow\ 0\! =\! \bar 0\! =\! \overline{f(w)}& =& \rm f(\overline w),\ \ i.e.\ \ w\ root\ of\ f\,\Rightarrow\, \overline w\ root\ of\ f\quad {\bf QED}
\end{eqnarray}$
This usually fails if $\rm\,f\,$ has nonreal coefficients, e.g. $\rm\,\bar w\,$ is a root of $\rm\,x\!-\!w\,$ iff $\rm\,\bar w = w,\,$ i.e. $\rm\,w\in \Bbb R.$
Remark $\ $ The analogous polynomial preservation property holds true for any algebraic structure, i.e. since homomorphisms preserve the basic operations (including constants = $0$-ary operations), they also preserve the "polynomial" terms composed of these basic operations. Said equivalently, homomorphisms commute with polynomials.
A: If the polynomial has real coefficients, then the answer is yes.
Indeed, let $\displaystyle p:=\sum_{i=0}^na_ix^i\in\mathbb{R}[x]$ and $z\in\mathbb{C}$ such that $p(z)=0$. Then, one has:
\begin{align}
p(\overline{z})&=\sum_{i=0}^na_i\overline{z}^i\\
&=\sum_{i=0}^na_i\overline{z^i}\\
&=\sum_{i=0}^n\overline{a_iz^i}\\
&=\overline{\sum_{i=0}^na_iz^i}\\
&=\overline{p(z)}\\
&=0
\end{align}
Therefore, $\overline{z}$ is a root of $p$.
However the result might be false if $p$ has complex coefficients.
Note that this fact can be generalized in the following fashion:

Let $L/K$ be a field extension, let $p\in K[x]$ and $z\in L$ such that $p(z)=0$. If $\sigma\colon L\rightarrow L$ is a ring homomorphism such that $\sigma_{\vert K}=\mathrm{id}_K$ i.e. $\sigma$ fixes the elements of $K$, then $\sigma(z)$ is a root of $p$.

A: Yes. Let $p(x) = a_nx^n+...+a_0$ and suppose $z$ is a root to $p(x)$. Then, $p(z)=0$. Let's consider $\bar{z}$.
$$
a_n\overline{z}^n+a_{n-1}\overline{z}^{n-1}+...+a_0.
$$
We know from the property of complex numbers that the above is equal to the following
$$
a_n\overline{z^n}+a_{n-1}\overline{z^{n-1}}+...+a_0.
$$
$$
=\overline{a_n{z}^n+a_{n-1}{z}^{n-1}+...+a_0.}
$$
$$
=\overline{p(z)}=\overline{0}=0
$$
Thus, $\overline{z}$ is a root to p(z) as well.
