Doubts about complex conjugate theorem This theorem states that a polynomial $p(x)$ with real coefficients has one roots in complex then other is its complex conjugate. Therefore any real number which is root of $p(x)$ can be written as $a+i0$ therefore $a-i0$ is also a root so it means it has the same root twice...
But it's not true in general... So I think I made a misunderstanding. Please help me to find out.
 A: For a proof: Suppose $z \in \mathbb{C}$ is a zero of the polynomial
$$f(x) = a_nx^n + a_{n-1}x^{n-1}+ \ldots + a_1x +a_0$$
where all $a_i \in \mathbb{R}$. Since $z$ is a zero of $f(x)$, we have that
$$0 = f(z)$$.
Consider the conjugate of $z$, denoted by $\overline{z}$, then we have that
$$f(\overline{z}) = a_n\overline{z}^n + a_{n-1}\overline{z}^{n-1}+ \ldots + a_1\overline{z} +a_0.$$
Note that we have the following properties:
$$\overline{z}^n = \overline{z} \cdot \overline{z} \cdot \ldots \cdot \overline{z} = \overline{z \cdot z \cdot \ldots \cdot z} = \overline{z^n}$$
and for $\lambda \in \mathbb{R}$
$$\lambda \overline{z} = \overline{\lambda \cdot z}$$
since $\overline{\lambda} = \lambda$. Moreover $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$.
This shows that 
$$a_n\overline{z}^n + a_{n-1}\overline{z}^{n-1}+ \ldots + a_1\overline{z} +a_0 = \overline{a_nz^n + a_{n-1}z^{n-1}+ \ldots + a_1z +a_0} = \overline{0} = 0.$$
Note that this proof really needs that the $a_i \in \mathbb{R}$, since we could only take the previous steps because $\overline{a_i} = a_i$ for $a_i \in \mathbb{R}$. 
As a counterexample for a polynomial which has complex coefficients: consider
$$p(z) = (z - i)(z + 3i) = z^2 + 2iz +3,$$ 
which has complex coefficients and although $i$ is a zero, $-i$ is not.
