Difference between revisions of "Complex conjugate root theorem"

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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.
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#REDIRECT[[Complex Conjugate Root Theorem]]
 
 
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 <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 <math>a_n, a_{n-1}, \ldots, a_1, a_0</math> are real numbers. 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 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 properties of complex conjugation,
 
<cmath>\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*}</cmath>
 
as required. <math>\square</math>
 
 
 
== See also ==
 
* [[Polynomial]]
 
 
 
[[Category:Algebra]]
 
[[Category:Polynomials]]
 
[[Category:Theorems]]
 

Latest revision as of 20:18, 27 November 2021