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It is specified as this:

$$(a,b)(c,d)=(ac-bd,ad+bc).\,$$

I don't see anywhere where the decision to have - these components $ac-bd$ and + these components $ad+bc$ was made. Or why it is like this. Wondering if one could explain why it's like this, without using or referencing $i$ or $\sqrt{-1}$, just using the idea of complex numbers as an ordered pair of reals.

Bill Dubuque
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user10869858
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  • Hm. If we treat $(a,b)$ as a vector of two elements (and of course the same for $(c,d)$), the left component of the multiplication looks something like the dot product, but with a subtraction instead. I wonder what that could mean. And if we were to concatenate the two into a $2 \times 2$ matrix columnwise like so:

    $$\begin{bmatrix} a & c\ b & d \end{bmatrix}$$

    then the right component of the multiplication is the determinant of this matrix but with addition in lieu of subtraction.

     – PrincessEev Mar 06 '19 at 09:10
  • Not sure if these mean anything because they're similar but not quite on the mark, but hopefully it means something. EDIT: Never mind, James posted an answer that filled in the details I was missing. – PrincessEev Mar 06 '19 at 09:10
  • "an ordered pair of vectors": you probably mean reals. –  Mar 06 '19 at 09:16
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    "Could someone explain why it is like this, without using or referencing the reason why it is like this?" ??? – hmakholm left over Monica Mar 06 '19 at 09:18
  • @HenningMakholm I am new to this so how would I know that that is the reason why it's like this lol. – user10869858 Mar 06 '19 at 09:24
  • I mention not using $i$ because that is what is done here. – user10869858 Mar 06 '19 at 09:29
  • @HenningMakholm I don't think ihat is the best choice of duplicate for this sort of question so I have voted to reopen – Bill Dubuque Mar 06 '19 at 21:35
  • @BillDubuque: The two questions seem to be asking fairly exactly the same. What makes you think they're not duplicates? – hmakholm left over Monica Mar 06 '19 at 22:11
  • @HenningMakholm There are many possible ways to answer this question, and the linked dupe has only one answer - which is far from representing all of the possible viewpoints. I think is is wrong to close with such a dupe target (even more so when it is a gold-badge closure with your own unique answer, when there are many possible answers) – Bill Dubuque Mar 06 '19 at 22:22
  • @BillDubuque: The previous instance of the question is not closed -- it is free for anyone to post more answers to it. – hmakholm left over Monica Mar 07 '19 at 10:52

4 Answers4

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This comes from the fact that every complex number $z=a+ib$ corresponds to a matrix of the form $$A=\begin{pmatrix}a & -b \\ b & a\end{pmatrix}.$$ Now take another matrix of this form, say $$B=\begin{pmatrix}c & -d \\ d & c\end{pmatrix}.$$ The product $$AB=\begin{pmatrix}a & -b \\ b & a\end{pmatrix}\begin{pmatrix}c & -d \\ d & c\end{pmatrix}=\begin{pmatrix}ac-bd & -(ad+bc) \\ (ad+bc) & ac-bd\end{pmatrix}.$$ You can again read of real and imaginary part of the product from this. If you are interested in why complex numbers correspond to such matrices, I can write a few lines about this fact here or you spend some minutes searching for it on the internet.

  • Why did this answer get a downvote? If you are nor satisfied, drop a comment to let me know what is wrong with it.. –  Mar 06 '19 at 09:15
  • Not me, I think this is helpful. Would be nice to include why matrices are related, though I have seen it mentioned and will look it up further. Having it in one place is nice. – user10869858 Mar 06 '19 at 09:16
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    This answer explains nothing. It just recasts the question in terms of matrices, and the "arbitrariness" of the definition remains. –  Mar 06 '19 at 09:18
  • @user10869858 There is a related question here: https://math.stackexchange.com/questions/1028371/complex-number-isomorphic-to-certain-2-times-2-matrices but I am willing to write a few lines about polar coordinates and the correspondence to matrices of this type, if you find nothing adequate –  Mar 06 '19 at 09:20
  • @YvesDaoust that's a good point, though knowing it's related to matrices at least adds some info to the mix. – user10869858 Mar 06 '19 at 09:20
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    @YvesDaoust I don't agree. I think it shows that the product comes from a matrix product and that is the reason to define it the way one does define it.. –  Mar 06 '19 at 09:21
  • @James: nonsense. In what way does that justify the $-$ and $+$ signs ? Read the question. –  Mar 06 '19 at 09:22
  • This plus and minus for real numbers. And as far as I read the question, it is asked why complex multiplication is defined the way it is. Because at first it looks very awkward. Thus understanding that it comes from matrix multiplication and is thus induced from something well-known is something that I consider helpful and also to answer why the definition is as it is. –  Mar 06 '19 at 09:24
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    The fact that you have a multiplication on $\mathbb R^2$ implies that any $(a,b)\in\mathbb R^2$ gives rise to a linear map $L_{(a,b)}:\mathbb R^2\to \mathbb R^2$ defined by $L_{(a,b)}(c,d)=(a,b)(c,d)=(ac-bd, ad+bc).$

    Making some plausible assumption, namely that $L_{(1,0)}$ is the identity matrix and that $L_{(0,1)}$ is given by rotation by $\pi/2$, and using the fact that $L_{(a,b)}(c,d)=L_{(c,d)}(a,b)$ for any $(a,b),(c,d)\in\mathbb R^2$, it is easy to check that the representing matrix of $L_{(a,b)}$ is the one appearing in James's answer.

     – kneidell Mar 06 '19 at 09:28
  • If anyone has a better answer, feel free and I will reevaluate, but for now this answers most of what I was looking for. Now I at least know it's based on matrix multiplication, would be nice to know why that was chosen as the multiplication method, but this is enough for now. Thank you. – user10869858 Mar 06 '19 at 09:28
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Your question is unreasonable, since it is because someone (Hamilton, in this case) was trying to formalize the complex numbers, as things of the form $a+bi$ with $a,b\in\mathbb R$ and $i^2=1$, that that definition was obtained.

José Carlos Santos
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    I disagree with the downvote. This is an excellent answer. The abstract formalism $(a,b)$ was derived from this. –  Mar 06 '19 at 09:15
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    This answer doesn't say anything to me, and it is a bit condescending. – user10869858 Mar 06 '19 at 09:18
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    @user10869858: not condescending at all. Just telling you the truth. Complex multiplication was not defined without referring to $i$. –  Mar 06 '19 at 09:20
  • Ah I see, makes sense. But it seems the Cayley Dickson construction doesn't reference $i$, maybe though in textbooks it does since it does ref complex numbers which are (typically I guess) defined by $i$. – user10869858 Mar 06 '19 at 09:23
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    No construction of $\mathbb C$ makes reference to $i$. The idea of constructing the complex numbers is that $i$ makes sense. – José Carlos Santos Mar 06 '19 at 09:25
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In the geometric interpretation of the complex numbers, $(a,b)$ is a point in the plane, the components of which are abscissa and ordinate.

Complex numbers are seen, in polar form, as a vector length and vector angle. The complex multiplication by $(0,1)$ is defined to be a rotation by a right angle. More generally, multiplication by a complex can be seen as the application of a similarity transform, i.e. the combination of a scaling and a rotation.

Now if you look at the product of two complex numbers as the combination of two similarity transforms, considering the rotation only, you need to enforce

$$(\cos\alpha,\sin\alpha)(\cos\beta,\sin\beta)=(\cos(\alpha+\beta),\sin(\alpha+\beta)),$$ as the combination of two rotations is a single rotation by the sum of the angles.

And recall,

$$(\cos(\alpha+\beta),\sin(\alpha+\beta))=(\cos\alpha\cos\beta\color{red}-\sin\alpha\sin\beta,\sin\alpha\cos\beta\color{red}+\cos\alpha\sin\beta).$$

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The elements of $\mathbb{R}^2$ with the element-wise addition and the multiplication defined above satisfies the field axioms (you can check it).

Botond
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    Okay, but that doesn't really enlighten anyone as to the inspiration underlying this definition. It forms a field with this definition, but I feel like OP is reaching more towards an idea of "why this definition?" – PrincessEev Mar 06 '19 at 09:15
  • @EeveeTrainer The inspiration is that we want to make a function $$, which will make $(\mathbb{R}^2, +, )$ a field. – Botond Mar 06 '19 at 09:20
  • I get that, but surely there are all sorts of definitions which would achieve that - which begs the question, why this particular one? – PrincessEev Mar 06 '19 at 09:21
  • @EeveeTrainer I haven't seen any other $*$ operation which would make a field with $\mathbb{R}^2$ and element-wise addition. Would you show me a different one? – Botond Mar 06 '19 at 09:23
  • I'll admit that I don't know of one off-hand, but we don't have the assurance that the multiplication is unique, either, so I would think that - even if I can't come up with one at this moment - there's at least a possibility another exists. You know what I mean? – PrincessEev Mar 06 '19 at 09:25
  • @EeveeTrainer You were right, we can make a lot of different field. An easy example that came in my mind is just to swap the "meanings" of the coordinates.I think we can restrict the possibilities with demanding $(a,0)(b,0)=(ab,0)$, but it won't solve every problem, right? – Botond Mar 06 '19 at 13:46