Complex Conjugate Root Theorem¶

Theorem¶

If $P$ is a polynomial in one variable with real coefficients, and $a + \mathrm{i} b$ is a root of $P$ with $a$ and $b$ being real numbers, then its complex conjugate $a - \mathrm{i} b$ is also a root of $P$ .

Interpretation¶

For a polynomial with real coefficients, the non-real roots come in pairs $a\pm\mathrm{i}b$ $i b$ : whenever $a+\mathrm{i}b$ is a root, so is its mirror image in the real axis.

Four roots of a real polynomial plotted on a grid in the complex plane, forming two conjugate pairs that are mirror images in the real axis.

The four plotted points are the roots of $p(z)=z^4+2z^3+2z^2+10z+25$ $25$ , a polynomial with real coefficients. They form the two conjugate pairs $1\pm2\mathrm{i}$ and $-2\pm\mathrm{i}$ $i$ , and each dashed segment joins a root to its mirror image. Each pair gives a real quadratic factor: $p(z)=(z^2-2z+5)(z^2+4z+5)$ $5)$ .

With real coefficients the roots are mirrored in the real axis. Give $a_0$ an imaginary part and the symmetry breaks: the theorem needs real coefficients.

The proof uses that conjugation respects sums and products, $\overline{z+w}=\overline{z}+\overline{w}$ and $\overline{zw}=\overline{z}\,\overline{w}$ $\overline{z} \overline{w}$ , and leaves real numbers unchanged. If $P(z)=c_nz^n+\cdots+c_1z+c_0$ with real $c_k$ and $P(\lambda)=0$ $0$ , then $P(\overline{\lambda})=\overline{P(\lambda)}=\overline{0}=0$ $0$ . The step $\overline{c_k}=c_k$ is where real coefficients are needed; without them the pairing can fail, as the polynomial $x-\mathrm{i}$ in the Examples shows.

For a non-real root $\lambda$ , the product $(z-\lambda)(z-\overline{\lambda})=z^2-2\operatorname{Re}(\lambda)z+|\lambda|^2$ is a real quadratic factor of $P$ , so non-real roots can be split off in pairs. Together with the fundamental theorem of algebra, this shows that every real polynomial is a product of real linear and real quadratic factors, and that a real polynomial of odd degree has a real root, as the Key Results state. The theorem returns later for eigenvalues: the non-real eigenvalues of a real matrix also come in conjugate pairs.

Key Results¶

If the degree of a real polynomial is odd, it must have at least one real root.
Any real $n\times n$ square matrix with $n$ odd (so its characteristic polynomial has odd degree $n$ ) has at least one real eigenvalue.

Examples¶

$x^2-2x+5$ has real coefficients and the root $1+2\mathrm{i}$ $2 i$ , so $1-2\mathrm{i}$ is also a root.
$x-\mathrm{i}$ has the root $\mathrm{i}$ but not $-\mathrm{i}$ : its coefficients are not all real, so the theorem does not apply.

Prerequisites¶

Complex Conjugate

Sources¶

Math II: Linear Algebra · Week 1

Aliases: conjugate root theorem, complex conjugate zeros theorem Symbols: lambda

Last updated: 20 October 2026 03:02