Difference between revisions of "Complex Conjugate Root Theorem"
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== Proof == | == 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> | + | 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*} | <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 + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0} = \overline{0} \\ | ||
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. 
