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
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $133$