Difference between revisions of "Complex Conjugate Root Theorem"
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| − | + | 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. | |
| + | |||
| + | A common intermediate step in intermediate competitions is to recognize that when given a complex root of a real polynomial, 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> for some real numbers <math>a_0, a_1, \ldots, a_n</math> and let <math>z</math> be a complex root of <math>P(x)</math>. We wish to show that <math>\overline{z}</math>, the complex conjugate of <math>z</math>, is also a root of <math>P(x)</math>. We have that <cmath>P(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 = 0.</cmath> Then by the [[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 also a root of <math>P(x)</math>, as required. <math>\square</math> | ||
| + | |||
| + | == See also == | ||
| + | * [[Polynomial]] | ||
| + | |||
| + | [[Category:Algebra]] | ||
| + | [[Category:Polynomials]] | ||
| + | [[Category:Theorems]] | ||
Latest revision as of 22:33, 1 April 2023
In algebra, the Complex Conjugate Root Theorem states that if 
is a polynomial with real coefficients, then a complex number is a root of if and only if its complex conjugate is also a root.
A common intermediate step in intermediate competitions is to recognize that when given a complex root of a real polynomial, its conjugate is also a root.
Proof
Let have the form 
for some real numbers 
and let 
be a complex root of . We wish to show that 
, the complex conjugate of , is also a root of . We have that ![\[P(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 = 0.\]](http://latex.artofproblemsolving.com/3/1/1/31170a751af29ffdb2bce97e99ae9c25e37bd639.png)
Then by the properties of complex conjugation,
which entails that is also a root of , as required. 
