Conjugate of Polynomial is Polynomial of Conjugate

Theorem
Let $f \left({z}\right) = a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0$ be a polynomial over complex numbers where $a_0, \ldots, a_n$ are real numbers.

Let $\alpha \in \C$ be a complex number.

Then:
 * $\overline {f \left({\alpha}\right)} = f \left({\overline \alpha}\right)$

where $\overline \alpha$ denotes the complex conjugate of $\alpha$.

Proof
By Power of Complex Conjugate is Complex Conjugate of Power:


 * $\overline {\alpha^k} = \left({\overline \alpha}\right)^k$

for all $k$ between $0$ and $n$.

Then from Product of Complex Conjugates:


 * $\overline {a_k \alpha^k} = \overline {a_k} \cdot \overline {\alpha^k}$

But $a_k$ is real.

So by Complex Number equals Conjugate iff Wholly Real:
 * $\overline {a_k} = a_k$.

From Sum of Complex Conjugates, it follows that:


 * $\overline {f \left({\alpha}\right)} = a_n \left({\overline \alpha}\right)^n + a_{n - 1} \left({\overline \alpha}\right)^{n - 1} + \cdots + a_1 \overline \alpha + a_0$

Hence the result.