Illustration of the matrix representations
The complex conjugation automorphism of C [of order 2] given by a + bi --> a - bi can be found back in the corresponding matrix model of C as transposition. For matrices in general, (A + B)T = AT + BT and (A*B)T = BT*AT [and (AT)T = A], but, since C is commutative, so is the model, and BT*AT = AT*BT. A more direct way to get the (composite) isomorphism
,
from C into M2(R), is by having matrices act from the right on row vectors. Then the matrix at hand is the representation matrix of the R-linear mapping of C given by z --> z*(a + bi) with respect to the basis {1, i} of C over R. We return to the more usual setting of left action on column vectors. Then the polar representation of complex numbers provides an even more obvious isomorphism:
.
We recognize the matrix of a positive [anti-clockwise] rotation around the origin over an angle of t radials. In particular, by the morphism property exp(i*(s + t)) = exp(is)*exp(it),
,
a rotation over an angle t followed by one over s comes down to a rotation over t + s, of course. As a consequence we regain the familiar sum formulas
,
[by the way, this is a particular instance of a general method to use group representations to derive identities between values of special functions]
Now, C has an outer automorphism of order 2 (complex conjugation, remember), so [inspired by the construction of cyclic extensions of groups] we shall search for a new element J, say, outside C, such that 
- J*(a + bi)*J -1 = a - bi,
- J2 = m, say, belongs to C
Since J obviously commutes with its square, the complex number m must actually be real and then may be scaled down to +1 or -1, for (J*z)2 = m*|z|2, assuming associativity etc.. So we are left with two cases m = 1 and m = -1. In either case we will seek an extension for the matrix model for C. In the first case we may take 
, in the second 
. The two respective extensions take the following form
