Exercise 826Algebra › Complex numbersHigh schoolAdded 3 Nov 2026

Conjugate roots of real polynomials

#conjugate roots#real polynomials#complex

Statement

Let $P$ be a polynomial with real coefficients. Show that if $z$ is a root of $P$ , then $\bar z$ is also a root.

Hint

Use $\overline{P(z)}=P(\bar z)$ for real coefficients.

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Coefficients. Write $P(X)=\sum a_kX^k$ with $a_k\in\mathbb{R}$ , so $\bar a_k=a_k$ .

Conjugation. For any $z$ :
$\overline{P(z)}=\overline{\sum a_kz^k}=\sum\bar a_k\bar z^k=\sum a_k\bar z^k=P(\bar z).$

Conclusion. If $P(z)=0$ , then $P(\bar z)=\overline{P(z)}=\bar 0=0$ . So $\bar z$ is also a root.

Consequence. Non-real roots of real polynomials come in conjugate pairs.