complex numbers
I have a question about imaginary numbers and their complex conjugate.
My teacher denotes the complex conjugate to have a bar over it.
For example:
$\overline{3+5i} = 3 – 5i$
But does the complex conjugate just switch the sign in-between the numbers?
For example, would
$\overline{-1 – i} = -1 + i$ ?
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 $v=\mathbf{0}$).
Every complex number can be written as $z=a+bi$, where $a,b\in \mathbb{R}$ (real numbers). The number $a$ is called real part of $z$ and the number $b$ is the imaginary part of $z$.
If the real part is zero then we call $z=bi$ as pure imaginary complex number.
Here is a diagram to show the inclusions:
Best Answer
It just switches the sign of the imaginary part of the number. Your example is correct.