Prove properties of polynomials of a complex variable with real coefficients

Consider the polynomial $f(z)$ of a complex variable with real coefficients.

Prove that
$\overline{f(z)} = f (\overline{z})$

for all $z \in \mathbb{C}$ .
Using part (a) show that if $f$ has any nonreal zeros, they must occur in pairs of a complex number and its conjugate.

Proof. Let
$f(z) = a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n \qquad a_i \in \mathbb{R}.$

Then, using the properties of conjugation we have,

$\begin{align*} \overline{f(z)} &= \overline{a_0 + a_1 z + \cdots + a_n z^n} \\ &= \overline{a_0} + \overline{a_1 z} + \cdots + \overline{a_n z^n} &(\text{since } \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}) \\ &= \overline{a_0} + \overline{a_1} \, \overline{z} + \cdots + \overline{a_n} \, \overline{z^n} &(\text{since } \overline{z_1 z_2} = \overline{z_1} \overline{z_2})\\ &= a_0 + a_1 \overline{z} + a_2 \overline{z^2} + \cdots + a_n \overline{z_n} &(\text{since } \overline{a} = a \text{ for } a \in \mathbb{R}) \\ &= a_0 + a_1 \overline{z} + a_2 (\overline{z})^2 + \cdots + a_n (\overline{z})^n \\ &= f(\overline{z}). \end{align*}$

(The final line follows since $\overline{z^2} = \overline{z \cdot z} = \overline{z} \cdot \overline{z} = (\overline{z})^2$ and then induction for to get $\overline{z^n} = (\overline{z})^n$ for all $n$ .) This completes the proof $. \qquad \blacksquare$
Proof. If $z$ is a non-real zero of $f$ then
$f(z) = 0 \quad \implies \quad \overline{f(z)} = 0 \quad \implies \quad f(\overline{z}) = 0,$

and $z \neq \overline{z}$ (since $z$ is not real by assumption). Hence, $\overline{z}$ is also a zero of $f$ , so the non-real zeros come in pairs $. \qquad \blacksquare$