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Say I have a complex polynomial:
$$a_0+a_1x+\cdots+a_nx^n,$$
where $a_0,\ldots,a_n$ are complex numbers.

What is the conjugate of this polynomial? How is it defined?

For example, if we have an inner product on the vector space $V$ defined by (for complex polynomials):
$$\int_0^1 p(x)\overline{q(x)}dx$$
Then how do we know that: $$\langle v,v \rangle = \int_0^1 (a_0+a_1x+\cdots+a_nx^n)(\overline{a_0}+\overline{a_1}x+\cdots+\overline{a_n}x^n)dx$$
is positive? Since there could be negative numbers in there, do we just know the positive ones outweigh the negative?

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  • $\begingroup$ It's the polynomial you get by replacing each $a_i$ with its complex conjugate. (And was $x_n$ meant to be $x^n$?) $\endgroup$ – Arturo Magidin Mar 7 '11 at 2:40
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It is simply the polynomial you get by replacing each $a_i$ by its complex conjugate.

You should think of this as the map induced on $\mathbb{C}[x]$ by the map $\mathbb{C}\to\mathbb{C}$ given by complex conjugation.

Added: How do we know that $\int_0^1 p(x)\overline{p(x)}\,dx$ is positive?

Write each $a_j = \alpha_j + i\beta_j$, with $\alpha_j,\beta_j\in\mathbb{R}$. Then note that $v = q(x)+ir(x)$, where \begin{align*} q(x) &= \alpha_0 + \alpha_1 x + \cdots + \alpha_nx^n\\ r(x) &= \beta_0 + \beta_1x + \cdots + \beta_nx^n. \end{align*} So you have: \begin{align*} \langle v,v\rangle &= \int_0^1 (a_0+a_1x + \cdots + a_nx^n)(\overline{a_0}+\overline{a_1}x + \cdots \overline{a_n}x^n)\,dx\\ &= \int_0^1(q(x)+ir(x))(q(x)-ir(x))\,dx\\ &= \int_0^1\Bigl( \left(q(x)\right)^2 - i^2\left(r(x)\right)^2\Bigr)\,dx\\ &= \int_0^1\Bigl( \left(q(x)\right)^2 + \left(r(x)\right)^2\Bigr)\,dx \end{align*} and since this is the integral of two nonnegative real valued functions, it is nonnegative (and equal to $0$ if and only if both $q(x)$ and $r(x)$ are zero, i.e., if and only if $v=\mathbf{0}$).

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    $\begingroup$ It seems that this is the most usual but not the only definition of "conjugate polynomial" that's in use. In a Google search, I found 3 texts that use this definition, but also this paper: citeseerx.ist.psu.edu/viewdoc/… that also reverses the order of the coefficients (see page 3). $\endgroup$ – joriki Mar 7 '11 at 4:01
  • $\begingroup$ @joriki: As with "normal", "conjugate" is somewhat overloaded, so it's not surprising to find it used to mean possibly different things... $\endgroup$ – Arturo Magidin Mar 7 '11 at 4:04
  • $\begingroup$ BTW, one of the texts gives a nice representation of the conjugate polynomial $\overline{Q}$ of a polynomial $Q$: $\overline{Q}(z) = \overline{Q(\overline{z})}$. $\endgroup$ – joriki Mar 7 '11 at 4:09
  • $\begingroup$ What's $p(x)$? Do you mean to say $q(x)+ir(x)$? $\endgroup$ – Justin Mar 7 '11 at 4:09
  • $\begingroup$ @Fdart17: Oops; $p(x)$ was indeed supposed to be $q(x)$. $\endgroup$ – Arturo Magidin Mar 7 '11 at 4:10

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