No. The standard field (that is, a ring where both operations are abelian groups) on the complex numbers doesn’t have a multiplicative inverse of 0; rings can’t have a multiplicative inverse for the additive identity. You can create an algebra with a ring as a sub-algebra with such, but it will no longer be a ring. My preferred method is to impose such an algebra on the one-point compactification of the Complex Numbers, where the single added point is denoted as “Ω”.
I started this project when I was 12, and when I could show that the results were self-consistent this was what I had settled on:
let z be a complex number that is not otherwise specified by the following equations. Note: the complex numbers contain the Real numbers, and so the following equations apply to the them as well.
0Ω=Ω0=1
z+Ω=Ω+z=zΩ=Ωz=Ω=ΩΩ
Ω-Ω=0. Ω-Ω=Ω+(-Ω)=Ω+(-1Ω)=Ω+Ω=0
The algebra described above is not associative. That is to say, (AB)C does not always equal A(BC).
No. The standard field (that is, a ring where both operations are abelian groups) on the complex numbers doesn’t have a multiplicative inverse of 0; rings can’t have a multiplicative inverse for the additive identity. You can create an algebra with a ring as a sub-algebra with such, but it will no longer be a ring. My preferred method is to impose such an algebra on the one-point compactification of the Complex Numbers, where the single added point is denoted as “Ω”.
I started this project when I was 12, and when I could show that the results were self-consistent this was what I had settled on:
let z be a complex number that is not otherwise specified by the following equations. Note: the complex numbers contain the Real numbers, and so the following equations apply to the them as well.
0Ω=Ω0=1
z+Ω=Ω+z=zΩ=Ωz=Ω=ΩΩ
Ω-Ω=0. Ω-Ω=Ω+(-Ω)=Ω+(-1Ω)=Ω+Ω=0
The algebra described above is not associative. That is to say, (AB)C does not always equal A(BC).