I'l like to drop in something else I have found (well, reconstructed from drips and drabs on the internet):

**Surcomplex Numbers**

The Surcomplex numbers are a proper superset (i.e., they contain) the Complex numbers. Also, 'surcomplex' is an awesome word.

Anyway, as you may or may not know, the complexes are formally defined as an ordered pair of reals, (r,i).

These ordered pairs are defined in set theory as {{r},{r,i}} (as are all ordered pairs). However, the arithmetic operations on them must be defined.

Addition:

(a,b)+(c,d)=(a+c,b+d)

Multiplication:

(a,b)*(c,d)=(ac-bd,bc+ad)

The inverses can be derived from these equations. As can exponentiation, etc.

Now, although the Surreals on their own cannot construct the square root of -1 (try it from the definition of multiplication I gave, you'll end up needing a set that can multiply with the empty set without becoming the empty set, which is impossible), neither can the reals. Bit using the same ordered pair methodology, but with Surreal numbers, a set of Surcomplex numbers can be defined:

(0,1)*(0,1)=(0*0-1*1,1*0+0*1)=(-1,0)

Any ordered pair with 0 as the right projection is a Surreal number, so (0,1) times itself in the Surcomplexes is -1. Just as in the Complexes, and so we shall label this number i.

Best thing about this, is we now have complex infinites and infinitesimals.

(ε,0)*(0,ω)=(ε*0-0*ω,0*0+εω)=(0,1)=i

(ε,4ω)*(1,2)=(ε-8ω,4ω+2ε)=(-8ω+ε,4ω+2ε)=(-8ω+ε)+(4ω+2ε)i

(0,ω)*(ω,0)=(0*ω-ω*0,ω*ω+0*0)=(0,ω^{2})=ω^{2}i

And, Seeing as the complexes remain unchanged as a subset of the Surcomplexes, e^{πi}+1=0. So, now all we need to do is find an identity that includes ω and ε as well!