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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.
| + | #REDIRECT[[Complex Conjugate Root Theorem]] |
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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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| − | == 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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| − | <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} + \overline{a_{n-1} z^{n-1}} + \cdots + \overline{a_1 z} + \overline{a_0} = 0 \\
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| − | a_n \overline{z^n} + a_{n-1} \overline{z^{n-1}} + \cdots + a_1 \overline{z} + a_0 = 0 \\
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| − | 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,
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| − | \end{align*}</cmath>
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| − | as required. <math>\square</math>
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| − | == See also ==
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| − | * [[Polynomial]]
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| − | [[Category:Algebra]]
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| − | [[Category:Polynomials]]
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| − | [[Category:Theorems]]
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