Complex Conjugate Root Theorem

Complex Conjugate Root Theorem states that for a real coefficient polynomial \(P(x)\), if \(a+bi\) (where \(i\) is the imaginary unit) is a root of \(P(x)\), then so is \(a-bi\)

To prove this, we need some lemma first.

Let \(x=a+bi,y=c+di \in \mathbb C\).
Let \(\overline \cdot\) denote the conjugate of a complex number.
\(\overline{x+y}=\overline{a+bi+c+di}=\overline{(a+c)+(b+d)i}=(a+c)-(b+d)i\)
\(\overline x + \overline y=\overline{a+bi} + \overline{c+di}=a-bi+c-di=(a+c)-(b+d)i\)
\(\square\)

Let \(x=a+bi,y=c+di \in \mathbb C\).
Let \(\overline \cdot\) denote the conjugate of a complex number.
\(\overline{xy}=\overline{(a+bi)(c+di)}=\overline{(ac-bd)+(bc+ad)i}=(ac-bd)-(bc+ad)i\)
\((\overline x) (\overline y)=(\overline{a+bi})(\overline{c+di})=(a-bi)(c-di)=(ac-bd)-(bc+ad)i\)
\(\square\)

This is immediate from Lemma \(2\).

Let \(\displaystyle P(x)=\sum_{i=0}^n a_i x^i\).
Let \(z\) be a root of \(P(x)\), i.e. \(P(z)=0\) (by factor theorem).
Then,
\(\;\;\;\;P(\overline z)\)
\(\displaystyle=\sum_{i=0}^n a_i \overline z^i\)
\(\displaystyle=\sum_{i=0}^n (\overline{a_i}) (\overline z^i)\) (real coefficients)
\(\displaystyle=\sum_{i=0}^n (\overline{a_i}) (\overline{z^i})\) (Lemma \(3\))
\(\displaystyle=\sum_{i=0}^n (\overline{a_i z^i})\) (Lemma \(2\))
\(\displaystyle=\overline{\sum_{i=0}^n a_i z^i}\) (Lemma \(1\))
\(=\overline{P(z)}\)
\(=\overline 0\)
\(=0\)
By factor theorem, \(\overline z\) is also a root.
\(\blacksquare\)