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We know complex numbers have a real and an imaginary part. We can visualize the imaginary part by adding another axis, hence a plane for both parts.

My question is really simple but may be unusual.

Why is the plus sign used to represent a complex number? $z=a+bi$. Is it just historically common to use the plus sign, or does one actually do arithmetic by using the sign? Why can't it be represented by a comma, for example, like a vector?

$z=a+bi$ versus $z=a,bi$

I'm really curious to know if $a+bi$ is a way of telling if the real and imaginary parts are connected or if it is a mathematical expression.

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    $\begingroup$ Ill counter your question with another question. Why do we write $1+3\sqrt{2}$? $\endgroup$ – Jonathan Davidson Jul 20 '17 at 21:52
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    $\begingroup$ Graphically, one really adds two vectors. $\endgroup$ – Bernard Jul 20 '17 at 21:52
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    $\begingroup$ You can define complex numbers as two-dimensional vectors, with $1 = (1,0)$ and $i = (0,1)$. Then $a + bi$ means $a(1) + b(i) = a(1,0) + b(0,1) = (a,0) + (0,b) = (a,b)$. $\endgroup$ – Jair Taylor Jul 20 '17 at 21:56
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    $\begingroup$ $a+bi$ comes from a history in algebra where that is the root of a polynomial. It was a couple of centuries and a bit of a conceptual leap before mathematicians picked up on the geometric implications. However, if it makes you feel any better, mathematicians frequently represent 2 dimensional vectors as $a\mathbf i + b\mathbf j$ so the notation is somewhat consistent. $\endgroup$ – Doug M Jul 20 '17 at 21:57
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    $\begingroup$ It's a good question and more subtle and ambiguous than one would think. You could most definitely use $\mathbb C = \mathbb R \times \mathbb R$ with the operation definitions $(a,b)+(c,d)= (a+c,b+d)$ and $(a,b)\cdot(c,d) = (ac - bd, bc+ad)$ and the plus wouldn't mean anything other than a place holder. But you can also think of $\mathbb C$ as an extension of the Reals with the ability to have roots to polynomials. In which case the plus acts very much as a plus. As $x = 5$ and $y=\sqrt{-1}$ would mean that $x + y$ is indeed $5 + i$. $\endgroup$ – fleablood Jul 20 '17 at 22:34
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The set of complex numbers has more structures than just being a set. One of the most important structures is that ${\bf C}$ is a field, which means one can do multiplication and division. It is very convenient to do these two operations with complex numbers represented as the form $a+bi$ instead of $(a,b)$ (or the worse form $(a,bi)$), the later of which would be rather cumbersome to use.

On the other hand, ${\bf C}$ is a vector space over ${\bf R}$ of dimension $2$. If one only cares about the (real) vector space structure of ${\bf C}$, one can indeed identify $a+bi\in{\bf C}$ as $(a,b)\in{\bf R}^2$ without sacrificing any clarity.


[Added later.] One of your comments above:

That was exactly the point of my question. To paraphrase, is the plus-symbol representing addition or is it not?

suggests that it would be helpful to go back to the very definition of complex numbers. One concrete and direct way to do it is considering the set $$ {\bf C}:=\{a+bi\mid a,b\in{\bf R}\}. $$ At this point, the plus sign "$+$" and the symbol "$i$" means nothing and similarly, $bi$ does not mean $b$ times $i$. One can see ${\bf C}$ as ${\bf R}^2$ in this definition.

The addition ($+_{\bf C}$), subtraction ($-_{\bf C}$), and multiplication operations($\cdot_{\bf C}$) can then be written down explicitly in these coordinates as

$$ (a+bi) +_{\bf C} (c+di) = (a+_{\bf R}c) + (b+_{\bf R}d)i \tag{1}$$

$$ (a+bi) -_{\bf C} (c+di) = (a-_{\bf R}c) + (b-_{\bf R}d)i \tag{2}$$

$$ (a+bi)\cdot_{\bf C} (c+di) = (ac-_{\bf R}bd) + (ad+_{\bf R}bc)i\tag{3}$$

Note that there are three different kinds of operations here

  • The known operations for ${\bf R}$: $+_{\bf R}$, $-_{\bf R}$, $\cdot_{\bf R}$ (we write $a\cdot_{\bf R}b$ as $ab$ above);
  • The operations we are defined for ${\bf C}$: $+_{\bf C}$, $-_{\bf C}$, $\cdot_{\bf C}$;
  • The formal symbol $+$ in the definition of the set ${\bf C}$.

We will show later that all these operations are compatible.

One can check that ${\bf C}$ together with the operations $+_{\bf C}$, $\cdot_{\bf C}$ forms a field. Further we identify $a\in{\bf R}$ as $a+0i$ so that ${\bf R}$ becomes a subset of ${\bf C}$. And we write $0+bi$ as $bi$ and $0+1i$ as $i$. Once we have this set up, we can check that for any $a,b\in{\bf R}$: $$ a+_{\bf C}[b\cdot_{\bf C}i]=(a+0i)+_{\bf C}[(b+0i)\cdot_{\bf C}(0+1i)] =(a+0i)+_{\bf C} (0+bi)=a+bi $$ which gives a meaning of the formal expression $a+bi$.

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  • $\begingroup$ Don't one loose one or more components of the dimension when using fields? $\endgroup$ – Natural Number Guy Jul 20 '17 at 23:03
  • $\begingroup$ In the Complex field. Or do you keep all the cartesian vector-components of the $R^2$ dimension? $\endgroup$ – Natural Number Guy Jul 20 '17 at 23:19
  • $\begingroup$ There is no such a thing called "$R^2$ dimension". ${\bf C}$ is a real vector space of dimension $2$ and a complex vector space of dimension $1$. $\endgroup$ – Jack Jul 21 '17 at 1:10
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    $\begingroup$ I am not convinced by this answer. How is $(a+bi)+(c+di)$ less cumbersome than $(a,b)+(c,d)$? $\endgroup$ – Federico Poloni Jul 21 '17 at 6:34
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    $\begingroup$ @FedericoPoloni: that's not what I say at all. The first paragraph is about "multiplication and division". $\endgroup$ – Jack Jul 21 '17 at 12:30
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I don't thing it answers your question but here is a funny consideration. I guess I saw it in a piece in some old issue of the "Kvant" magazine (a popular science magazine in physics and mathematics, pretty famous in the Soviet Union; see https://en.wikipedia.org/wiki/Kvant_(magazine)). Later I learned that the idea for this piece was pitched to the authors by I.M.Gelfand.

The rule for addition of the fractions is way too complicated. A lazy pupil don't like it, so he tries to do it in the easiest way: $$\frac{a}{b} + \frac{c}{d} = \frac{a+c}{b+d}.$$

The problem is that then one loses the distributivity of multiplication. After a bit of playing with formulas the lazy pupil finds a nice way to multiplicate them:

$$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac - bd}{ac + bd}.$$

I think there is no need to explain what happens next. Then he starts to investigate properties of the resulting algebraic structure, and to my opinion this is one of the most exciting introductions to the complex numbers I've ever seen.

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We use the + sign because $i$ is just another number, and we really are adding the imaginary and real parts together. However, unlike "$1 + 2$" where we have a nice tidy alternate way of writing the result, "$3$", with "$1 + 2i$", there is no more convenient expression for it, so we leave it as $1 + 2i$.

And $1,2i$ is definitely a less convenient expression than $1 + 2i$. One simple example, it is obvious that $3(1 + 2i)$ should be $3\cdot 1 + 3\cdot 2i = 3 + 6i$. This is just distributivity of multiplication over addition. But if you use $3(1,2i)$, then you also have to establish that multiplication also distributes over ",". If "," is just going to mean the same thing as "+" already does, why do we need to introduce it and new rules to go with it?

You may have had complex numbers introduced to you as ordered pairs (this is how they were introduced to me), but this is not really the case. Ordered pairs provide a model for complex numbers - i.e., a concrete set with operations that behave exactly like complex numbers should. But it is more convenient, and more intuitive, to consider complex numbers more abstractly - to think of them just as "what do you get if you add a new number $i$ to the reals with the property that $i^2 = -1$", without trying to define some internal structure for $i$. All that is needed is that it is some object that is not any of the Real numbers. If you then insist that it should be addable to and multipliable by the real numbers, and demand that all the normal rules of multiplication and addition still hold, the full complex numbers is what you get.

Historically, this is how complex numbers were first developed. They were introduced in the 16th century to solve cubic equations. Such equations were known to always have at least one solution, but that solution could not always be found. A known method for solving some cubics involved first solving a related quadratic, then taking the cube roots of the two solutions and adding them together to get a solution to the cubic. However, the method failed when the discriminant of the quadratic was negative.

A fellow named Scipione del Ferro decided to pretend the quadratic with negative discriminant had solutions anyway. Assuming that these fake numbers still behaved like all the others, he proceeded with the rest of the previous method, and discovered that adding the two cube roots caused the fake numbers to cancel out, leaving a "real" number only. That number he confirmed actually did solve the original cubic. Once this became known (an intriguing bit of history involving mathematics duels and broken promises) people accepted that if these "imaginary" numbers somehow managed to produce useful results, they must exist after all.

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  • $\begingroup$ Maybe the confusing bit for me is that "plus" in the natural sense means that there is a concrete measurable (existing) number on both sides of the plus-sign, so that it always gives a natural or real solution (like what we learned about counting in school), while in a complex number this is not always the case? That (most or many) solutions are undefined. Like you pointed out on cubic equations some leaving the real or natural number only as solution. Is this proof or disproof that there exist a real or natural computation in their sets for all complex equations that has one unique solution? $\endgroup$ – Natural Number Guy Jul 21 '17 at 1:17
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    $\begingroup$ Other than the name, there is nor eason to think imaginary numbers are not as "real" as the real numbers, and could not be measured. Physicists can and do measure imaginary numbers regularly. The fact that you can't simplify 1+2i further is not different to the fact that you cannot simplify 1+2**0.5 further. So the "+" is just as much an addition of normal numbers as it is without imaginary numbers. So in the end, it's just the name that is confusing, but this is the case for "negative" numbers and "irrational" numbers as well, which gave people the same uneasy feelings when introduced. $\endgroup$ – Marc Lehmann Jul 21 '17 at 6:51
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    $\begingroup$ @MarcLehmann what imaginary numbers do physicists measure? (in my field of electronics, although you can represent phase and magnitude by a single complex number, but to obtain phase and magnitude you make multiple measurements of the real value of the signal, so you only ever measure real values, not imaginary numbers) $\endgroup$ – Pete Kirkham Jul 21 '17 at 13:01
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A lot of things have been said in the comments already, but let me try to put things into context:

The complex numbers are defined in a way that they behave just as you would expect algebraic expressions of the form $a+ib$ to behave, e.g. under addition and multiplication (so you have associativity, commutativity, distributivity..) with the addition that you then have to define what the product of two imaginary numbers is, which is done via defining $i^2 = - 1$.

You could in principle invent a new symbol to denote complex numbers, but the "+" notation has a lot of advantages in that it is intuitive (as pointed out above) and can be considered as a vector-valued statement. So you can write $a + ib = a\cdot 1 + b \cdot i$ and consider the symbols $1$ and $i$ as the two vectors $(1,0)$ and $(0,1)$, so you add another dimension to the real line to get the complex plane. So as a set, they are indistinguishable from $\mathbb{R}^{2}$.

As Jack, who was faster than me, pointed out, the complex numbers have a lot more structure than just being a set. They are, for example, a vector space over the field of real numbers, $\mathbb{R}$, since linear combinations with real coefficients of complex numbers are again complex. And if you consider it as such a space, they are actually isomorphic to (and hence indistinguishable from) $\mathbb{R}^{2}$.

The complex numbers can also be considered as a metric space (and hence a topological space) if you define as their metric the distance function in the complex plane, i.e. $$ d(z_{1}, z_{2}) := | z_{1} - z_{2} |. $$ Then, even as a metric space, they are indistinguishable from $\mathbb{R}^{2}$ (with the Euclidean metric). This means that you have the same kinds of open sets, the same notion of convergence (i.e. what it means for a sequence $(z_{n})_{n \in \mathbb{N}}$ to converge to a limit $z$) etc.

There are even more structures present here, for example $\mathbb{C}$ is also a normed space (because its metric is actually defined via a norm).

But most importantly, and this is what makes $\mathbb{C}$ so special and sets it apart from $\mathbb{R}^{2}$: they are a field. So you can not only add complex numbers (as you can with any vectors), but you can also multiply them in a natural way. And this multiplication has very far reaching consequences. It is, for example, the reason, why complex differentiability is so much stronger than total differentiability in $\mathbb{R}^{2}$.

So what I - and others - want to say is that, for many practical purposes, it is very convenient (and justified) to consider a complex number as the sum of its real and imaginary part, as if they were the components of a two-dimensional vector. But complex numbers are much more than just vectors, which is what makes them so special - and so interesting.

Hope that helps!

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  • $\begingroup$ First two paragraphs helped. The rest is a bit too higher level for me. I think I am more interested in the $R^2$ space than the Complex one. Im trying to connect or understand the two. Seems that it actually is components in the $R^2$-dimensional space. $\endgroup$ – Natural Number Guy Jul 20 '17 at 22:40
  • $\begingroup$ Yes, for most purposes you can consider it that way, unless you want to multiply complex numbers - there it gets really different from $\mathbb{R}^{2}$ because you cannot "really" multiply vectors with each other. There is, of course, the possibility to take the scalar product $\vec{v} \cdot \vec{w}$, but this returns a number, not a vector. Whereas the complex multiplication gives back another complex number. $\endgroup$ – Andre Jul 20 '17 at 22:43
  • $\begingroup$ You can also take the cross product, but then you have three dimensions to worry about. $\endgroup$ – Kevin Jul 21 '17 at 0:59
  • $\begingroup$ @Kevin or you can take the wedge product, but the wedge product is not the most obvious multiplication rule for vectors... $\endgroup$ – Brevan Ellefsen Jul 21 '17 at 6:19
  • $\begingroup$ Well, the cross product is identical to the wedge product in 3 dimensions. And in any case, this is not a "good" multiplication in that it is not commutative and does not have a unit element, so yes, you can consider this a multiplication, but it is very different from the multiplication of the field $\mathbb{R}$. $\endgroup$ – Andre Jul 21 '17 at 13:26
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I liken the plus sign as a means to distinguish unlike terms. If one were to write 2 + 3 instead of 2 + 3i, the first simplifies to 5, but the second doesn't.

My preference is to use ordered pairs (x, y) for elements of C, but then to explain the shortcut notation x + yi or x + iy. Specifically, it exploits our knowledge of arithmetic and algebra with real numbers, and particularly how to multiply.

So, my answer is that the plus sign distinguishes so-called real and imaginary parts from each other. I think ordered pairs do this better, but one should then add in the x + yi to take advantage of the familiarity.

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