|
|
| Line 1: |
Line 1: |
| − | In [[algebra]], the '''complex conjugate root theorem''' states that if <math>P(x)</math> is a [[polynomial]] with [[real number | real coefficients]], then a [[complex number]] is a root of <math>P(x)</math> if and only if its [[complex conjugate]] is also a root.
| + | #REDIRECT[[Complex Conjugate Root Theorem]] |
| − | | |
| − | A common intermediate step is to present 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 <math>P(x)</math> have the form <math>a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0</math>, where constants <math>a_n, a_{n-1}, \ldots, a_1, a_0</math> are real numbers, and let <math>z</math> be a complex root of <math>P(x)</math>. We then wish to show that <math>\overline{z}</math>, the complex conjugate of <math>z</math>, is also a root of <math>P(x)</math>. Because <math>z</math> is a root of <math>P(x)</math>, <cmath>P(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 = 0.</cmath> Then by the [https://artofproblemsolving.com/wiki/index.php/Complex_conjugate#Properties properties of complex conjugation],
| |
| − | <cmath>\begin{align*}
| |
| − | \overline{a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0} = \overline{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*}</cmath>
| |
| − | which entails that <math>\overline{z}</math> is a root of <math>P(x)</math>, as required. <math>\square</math>
| |
| − | | |
| − | == See also ==
| |
| − | * [[Polynomial]]
| |
| − | | |
| − | [[Category:Algebra]]
| |
| − | [[Category:Polynomials]]
| |
| − | [[Category:Theorems]] | |
