Difference between revisions of "Complex conjugate root theorem"

Revision as of 12:06, 14 July 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 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 $P(x)$ 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 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, $\begin{align*} \overline{a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0} = 0 \\ \overline{a_n z^n} + \overline{a_{n-1} z^{n-1}} + \cdots + \overline{a_1 z} + \overline{a_0} = 0 \\ a_n \overline{z^n} + a_{n-1} \overline{z^{n-1}} + \cdots + a_1 \overline{z} + a_0 = 0 \\ a_n \overline{z}^n + a_{n-1} \overline{z}^{n-1} + \cdots + a_1 \overline{z} + a_0 = 0 \\ P(\overline{z}) = 0, \end{align*}$ as required. $\square$

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 [[Category:Algebra]]
-[[Category:Polynomial]]
+[[Category:Polynomials]]
 [[Category:Theorems]]

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Difference between revisions of "Complex conjugate root theorem"

Revision as of 12:06, 14 July 2021

Proof

See also