# 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.

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## Lemma \(1\) --- Conjugate of Sum is Sum of Conjugate

Let \(x=a+bi,y=c+di \in \mathbb C\).

Let \(\overline \cdot\) denote the conjugate of a complex number.

\(\overline{x+y}=\overline{a+bi+c+di}=\overline{(a+c)+(b+d)i}=(a+c)-(b+d)i\)

\(\overline x + \overline y=\overline{a+bi} + \overline{c+di}=a-bi+c-di=(a+c)-(b+d)i\)

\(\square\)

## Lemma \(2\) --- Conjugate of Product is Product of Conjugate

Let \(x=a+bi,y=c+di \in \mathbb C\).

Let \(\overline \cdot\) denote the conjugate of a complex number.

\(\overline{xy}=\overline{(a+bi)(c+di)}=\overline{(ac-bd)+(bc+ad)i}=(ac-bd)-(bc+ad)i\)

\((\overline x) (\overline y)=(\overline{a+bi})(\overline{c+di})=(a-bi)(c-di)=(ac-bd)-(bc+ad)i\)

\(\square\)

## Lemma \(3\) --- Conjugate of Power is Power of Conjugate

This is immediate from Lemma \(2\).

## Proof of the Theorem

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.

\(\blacksquare\)

**Cite as:**Complex Conjugate Root Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/complex-conjugate-root-theorem/