Quick question about notation (Linear Algebra)

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p\in P_3(\mathbb{F})

What does \overline{p(z)} mean?

I would guess that it's related to the complex conjugate, but I'm not sure. For context, I'm dealing with an inner product space defined by \langle p,q\rangle=\intop_{0}^{1}p(z)\overline{q(z)}dz

Thanks!
 
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Yes, it's a complex conjugate.
 
that's an "inner product" of complex valued functions. The reason for the complex conjugate is so that
<f(z),f(z)>= \int_0^1 f(z)\overline{f(z)}dz
will be a non-negative real number.
 
Inner Product Space Norm

Thanks for the help so far. I'm trying to find the norm for the this inner product. So far I have:

\Vert p\Vert=\sqrt{\left\langle p,p\right\rangle }=\sqrt{\intop_{0}^{1}p(z)\overline{p(z)}dz}=\sqrt{\intop_{0}^{1}|p(z)|^{2}dz}

I also know that p(z) can be denoted \sum_{i=0}^{n}a_{i}z^{i} but I don't know how to connect the two beyond placing the sum within the absolute value.

Is there a way to simplify |\sum_{i=0}^{n}a_{i}z^{i}|^{2}?
 
You've already written the norm of the inner product. Do you want to write it in terms of the a_i's? Is P_3 third degree polynomials? Or second? Either way it's going to be kind of uselessly complicated. If it's second, e.g. write (a0+a1*z+a2*z^2) times the complex conjugate and integrate over the real variable z in [0,1].
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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