Conjugate of Real Polynomial is Polynomial in Conjugate

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Theorem

Let $\map P z$ be a polynomial in a complex number $z$.

Let the coefficients of $P$ all be real.


Then:

$\overline {\map P z} = \map P {\overline z}$

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


Proof

Let $\map P z$ be expressed as:

$a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0$

Then:

\(\ds \overline {\map P z}\) \(=\) \(\ds \overline {a_n z^n + a_{n-1} z^{n - 1} + \cdots + a_1 z + a_0}\)
\(\ds \) \(=\) \(\ds \overline {a_n z^n} + \overline {a_{n - 1} z^{n - 1} } + \cdots + \overline {a_1 z} + \overline {a_0}\) Sum of Complex Conjugates
\(\ds \) \(=\) \(\ds \overline {a_n} \overline {z^n} + \overline {a_{n - 1} } \overline {z^{n - 1} } + \cdots + \overline {a_1} \overline z + \overline {a_0}\) Product of Complex Conjugates
\(\ds \) \(=\) \(\ds a_n \overline {z^n} + a_{n - 1} \overline {z^{n - 1} } + \cdots + a_1 \overline z + a_0\) Complex Number equals Conjugate iff Wholly Real
\(\ds \) \(=\) \(\ds a_n \overline z^n + a_{n - 1} \overline z^{n - 1} + \cdots + a_1 \overline z + a_0\) Product of Complex Conjugates/General Result
\(\ds \) \(=\) \(\ds \map P {\overline z}\)

$\blacksquare$


Sources