# Representing the modulus and complex conjugate in matrix terms

• Dec 8th 2010, 06:20 AM
juanma101285
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
• Dec 8th 2010, 07:07 AM
FernandoRevilla
Denote:

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

Then, the map

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

is bijective and satisfies:

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

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

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

So,

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

etc.

Regards.

Fernando Revilla
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