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Math Help - Representing the modulus and complex conjugate in matrix terms

  1. #1
    Junior Member
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    Representing the modulus and complex conjugate in matrix terms

    Hi, I have the following two questions...I am not sure if I am in the right direction with the first one and I have no clue about the second one :-/

    1.- Consider the set C of all matrices (with real entries) of the form

    (sorry, I don't know how to code matrices! I'll separate each element with "|")

    (a | -b)
    (b | a)

    The set C of matrices can clearly be identified with the complex numbers.

    So I am a bit lost here... Why? Is it because, for example, if z=a+ib, its complex conjugate will be a-ib?


    The next question is...

    Continuing the previous exercise, how do the modulus and complex conjugate appear when translated into matrix terms? How does the reciprocal of the complex number z appears?


    I would really appreciate it if you could give me a hint, thanks! I also tried to write the matrices using the code, but I go it wrong... I do not know if I am using an old version of the code or something
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Denote:

    <br />
\mathcal{C}=\left\{{A(a,b):\;a,b\in{\mathbb{R}}}\r  ight\},\quad A(a,b)=\begin{bmatrix}{a}&{-b}\\{b}&{\;\;a}\end{bmatrix}

    Then, the map

    f:\mathbb{C}\rightarrow{\mathcal{C}},\quad f(a+bi)=A(a,b)

    is bijective and satisfies:

    (i) f[(a+bi)+(a'+b'i)]=f(a+bi)+f(a'+b'i)

    (ii) f[(a+bi)\cdot (a'+b'i)]=f(a+bi)\cdot f(a'+b'i)

    this means that f is an isomorphism and translates the structure of field from \mathbb{C} to \mathcal{C}.

    So,

    |a+bi|^2=a^2+b^2=\det (A(a,b))

    etc.

    Regards.

    Fernando Revilla
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