In mathematics, the complex conjugate root theorem states that if P is a polynomial in one variable with real coefficients, and a + bi is a root of P with a and b real numbers, then its complex conjugate a − bi is also a root of P.^[1]

It follows from this (and the fundamental theorem of algebra) that, if the degree of a real polynomial is odd, it must have at least one real root.^[2] That fact can also be proved by using the intermediate value theorem.

Examples and consequences

• The polynomial x² + 1 = 0 has roots ± i.
• Any real square matrix of odd degree has at least one real eigenvalue. For example, if the matrix is orthogonal, then 1 or −1 is an eigenvalue.
• The polynomial

${\displaystyle x^{3}-7x^{2}+41x-87}$

has roots

${\displaystyle 3,\,2+5i,\,2-5i,}$

and thus can be factored as

${\displaystyle (x-3)(x-2-5i)(x-2+5i).}$

In computing the product of the last two factors, the imaginary parts cancel, and we get

${\displaystyle (x-3)(x^{2}-4x+29).}$

The non-real factors come in pairs which when multiplied give quadratic polynomials with real coefficients. Since every polynomial with complex coefficients can be factored into 1st-degree factors (that is one way of stating the fundamental theorem of algebra), it follows that every polynomial with real coefficients can be factored into factors of degree no higher than 2: just 1st-degree and quadratic factors.

• If the roots are a+bi and a−bi, they form a quadratic

${\displaystyle x^{2}-2ax+(a^{2}+b^{2})}$.

If the third root is c, this becomes

${\displaystyle (x^{2}-2ax+(a^{2}+b^{2}))(x-c)}$

${\displaystyle =x^{3}+x^{2}(-2a-c)+x(2ac+a^{2}+b^{2})-c(a^{2}+b^{2})}$.

Proof

One proof of the theorem is as follows:^[2]

Consider the polynomial

${\displaystyle P(z)=a_{0}+a_{1}z+a_{2}z^{2}+\cdots +a_{n}z^{n}}$

where all a_r are real. Suppose some complex number ζ is a root of P, that is ${\displaystyle P(\zeta )=0}$. It needs to be shown that

${\displaystyle P{\big (}\,{\overline {\zeta }}\,{\big )}=0}$

as well.

If P(ζ  ) = 0, then

${\displaystyle a_{0}+a_{1}\zeta +a_{2}\zeta ^{2}+\cdots +a_{n}\zeta ^{n}=0}$

which can be put as

${\displaystyle \sum _{r=0}^{n}a_{r}\zeta ^{r}=0.}$

Now

${\displaystyle P{\big (}\,{\overline {\zeta }}\,{\big )}=\sum _{r=0}^{n}a_{r}{\big (}\,{\overline {\zeta }}\,{\big )}^{r}}$

and given the properties of complex conjugation,

${\displaystyle \sum _{r=0}^{n}a_{r}{\big (}\,{\overline {\zeta }}\,{\big )}^{r}=\sum _{r=0}^{n}a_{r}{\overline {\zeta ^{r}}}=\sum _{r=0}^{n}{\overline {a_{r}\zeta ^{r}}}={\overline {\sum _{r=0}^{n}a_{r}\zeta ^{r}}}.}$

Since

${\displaystyle {\overline {\sum _{r=0}^{n}a_{r}\zeta ^{r}}}={\overline {0}},}$

it follows that

${\displaystyle \sum _{r=0}^{n}a_{r}{\big (}\,{\overline {\zeta }}\,{\big )}^{r}={\overline {0}}=0.}$

That is,

${\displaystyle P{\big (}\,{\overline {\zeta }}\,{\big )}=a_{0}+a_{1}{\overline {\zeta }}+a_{2}{\big (}\,{\overline {\zeta }}\,{\big )}^{2}+\cdots +a_{n}{\big (}\,{\overline {\zeta }}\,{\big )}^{n}=0.}$

Note that this works only because the a_r are real, that is, ${\displaystyle {\overline {a_{r}}}=a_{r}}$. If any of the coefficients were non-real, the roots would not necessarily come in conjugate pairs.

Notes