There are two different ways to represent a complex number with $2 \times 2$ real matrices: $$ \rho: \mathbb{C} \rightarrow M_2(\mathbb{R}) \qquad \rho(z)=\rho(a+ib)= \left[ \begin{array}{ccccc} a&b \\ -b &a \end{array} \right] $$ and $$ \bar\rho: \mathbb{C} \rightarrow M_2(\mathbb{R}) \qquad \bar\rho(z)=\bar\rho(a+ib)= \left[ \begin{array}{ccccc} a&-b \\ b &a \end{array} \right] $$ We can find one or the other in many sources, and somewhere both, as in History of the matrix representation of complex numbers

Clearly this two representations generate the same subring of $M_2(\mathbb{R})$ but ve have: $\bar \rho(z)=\rho(\bar z)$.

The existence of this double representation has some (hidden for me) significance or is purely fortuitous ?

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    $\begingroup$ There are actually lots of ways: for any matrix $A$ satisfying $A^2 = -I$, the subring generated by $I$ and $A$ is isomorphic to the complex numbers. Although, I think the one you list is the only one where complex conjugation is given by the transpose. $\endgroup$ – Hurkyl Dec 31 '14 at 18:35

There’s an arbitrariness in putting $i$ above the real axis instead of below. This is the same.

The complex numbers have a fundamental analytical and algebraic symmetry given by the complex conjugation. Algebraically, this just reflects the arbitrary choice of one of the roots of $X^2 + 1$ and expressing the other one in terms of the chosen one.


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