Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Complex conjugate root theorem"
Line 1: Line 1:
−
The '''complex conjugate root theorem''' states that if <math>P(x)</math> is a [[polynomial]] with [[real number | real coefficents]], then a complex number is a root of a polynomial if and only if its [[complex conjugate]] is also a root.
+
The '''complex conjugate root theorem''' states that if <math>P(x)</math> is a [[polynomial]] with [[real number | real coefficents]], then a [[complex number]] is a root of <math>P(x)</math> if and only if its [[complex conjugate]] is also a root.
  
 
A common setup in contest math is giving a complex root of a real polynomial without its conjugate. It is then up to the solver to recognize that its conjugate is also a root.
 
A common setup in contest math is giving a complex root of a real polynomial without its conjugate. It is then up to the solver to recognize that its conjugate is also a root.

Revision as of 11:56, 14 July 2021

The complex conjugate root theorem states that if $P(x)$ is a polynomial with real coefficents, then a complex number is a root of $P(x)$ if and only if its complex conjugate is also a root.

A common setup in contest math is giving a complex root of a real polynomial without its conjugate. It is then up to the solver to recognize that its conjugate is also a root.

Proof

Let $P(x)$ have the form $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$, where $a_n, a_{n-1}, \ldots, a_1, a_0$ are real numbers. Let $z$ be a complex root of $P(x)$. We then wish to show that $\overline{z}$, the complex conjugate of $z$ is a root of $P(x)$. Because $z$ is a root of $P(x)$, \[P(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 = 0.\] Then by the properties of complex conjugation, \begin{align*} \overline{a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0} = 0 \\ \overline{a_n z^n} + \overline{a_{n-1} z^{n-1}} + \cdots + \overline{a_1 z} + \overline{a_0} = 0 \\ a_n \overline{z^n} + a_{n-1} \overline{z^{n-1}} + \cdots + a_1 \overline{z} + a_0 = 0 \\ a_n \overline{z}^n + a_{n-1} \overline{z}^{n-1} + \cdots + a_1 \overline{z} + a_0 = 0 \\ P(\overline{z}) = 0, \end{align*} as required. $\square$

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Complex_conjugate_root_theorem&oldid=158419"