# Complex numbers with conjugate multiplication - field or …?

I could have sworn that when we learned about complex numbers in signals and systems that they form a field in (at least) two ways, depending on multiplication, which is most intuitively described in polar coordinates:

Normal multiplication adds the arguments' phases, while conjugate multiplication subtracts them.

But, whereas (scalar) phase addition is associative, subtraction is only left associative.

So what algeraic structure does $\mathbb C$ under complex conjugation form?

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If I use $z\times w$ to mean conjugate multiplication, and $zw$ to mean ordinary multiplication, you are talking about $z\times w=z\overline w$, right? – Jonas Meyer Feb 26 '13 at 5:15
Yes $zw$ versus $z \overline w$ – alancalvitti Feb 26 '13 at 5:16
It does not subtract them, it adds their opposites. Unless you actually talk about $z\cdot \bar{w}$. In which case your are looking at an inner product space. – 1015 Feb 26 '13 at 5:17
@julien, can you explain what you mean by 'opposites'? conjugate multiplication is not commutative - it only take the complex conjugate of the 2nd argument. – alancalvitti Feb 26 '13 at 5:19
I'm lost. I simply mean: if $z=re^{i\theta}$ and $w=\rho e^{i\phi}$, then $\overline{zw}=r\rho e^{-i(\theta+\phi)}$. But I'm afraid I misunderstood something. I think you talk about $z\cdot\bar{w}$, and not $\overline{zw}$. – 1015 Feb 26 '13 at 5:21

The map $\mathbb C\times \mathbb C\to\mathbb C$ defined by $(z,w)\mapsto z\overline w$ is the standard inner product on $\mathbb C$. Thought of in this way, it isn't a multiplication in a ring, but rather it is thought of in the same way that an inner product on any other complex vector space $V$ defines a map $V\times V\to \mathbb C$ with certain properties.

But if we think of $\mathbb C$ together with "conjugate multiplication" and ordinary addition as operations on $\mathbb C$, then multiplication is nonassociative (and noncommutative), but still distributive over addition, hence $\mathbb C$ with this structure is a nonassociative ring. It still has some nice properties, like a right (but no left, thanks Hurkyl) multiplicative identity, multiplicative inverses for all nonzero elements, and cancellation.

I don't know if it falls under a particularly well studied class of nonassociative rings, but there are books written on nonassociative rings and algebras where you might find something. Like $\mathbb C$ with its ordinary multiplication, the complex conjugation operation $\mathbb C\to\mathbb C$ still has an important role, with its relation to multiplication given by $\overline{z\times w}=\overline w\times \overline z$.

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I do believe you only have a right identity. – Hurkyl Feb 26 '13 at 5:57
You're right, thanks for catching my careless error. – Jonas Meyer Feb 26 '13 at 5:58
But now that I've said that, there is a weird sort of left identity formed by left-multiplying by 1, and then again by 1. – Hurkyl Feb 26 '13 at 5:59
@Hurkyl: Yes, or generally $z\times (1\times w)=zw$. Interesting that conjugation becomes left multiplication by $1$. – Jonas Meyer Feb 26 '13 at 6:05

A *-ring, or ring with involution. Important examples include C*-algebras.

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I do not think that alancalvitti is asking about $\mathbb C$ together with its ordinary addition, multiplication, and separate conjugation, but rather $\mathbb C$ with ordinary addition and "conjugate multiplication" $(z,w)\mapsto z\overline w$. I mention this because usually rings are assumed associative, but it is a nonassociative $*$-ring. – Jonas Meyer Feb 26 '13 at 5:31
Jonas is right, that's what I'm asking. – alancalvitti Apr 6 '13 at 19:12