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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 <math>P(x)</math> 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.
  
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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.
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A common setup in contest math is presenting 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 ==
 
== Proof ==
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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,  
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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 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*}
 
<cmath>\begin{align*}
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\overline{a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0} = 0 \\
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\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 \\
 
\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 \\
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P(\overline{z}) = 0,
 
P(\overline{z}) = 0,
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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as required. <math>\square</math>
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which entails that <math>\overline{z}</math> is a root of <math>P(x)</math>, as required. <math>\square</math>
  
 
== See also ==
 
== See also ==

Revision as of 18:48, 21 August 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 presenting 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 also 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} = \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*} which entails that $\overline{z}$ is a root of $P(x)$, as required. $\square$

See also

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