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The field $\mathbb{C} $ of complex numbers is well-defined by the Hamilton axioms of addition and product between complex numbers, i.e., a complex number $z$ is a ordered pair of real numbers $(x,y)$, which satisfy the following operations $+$ and $\cdot$:

$ (x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1+x_2) $

$(x_1,y_1)(x_2,y_2) = (x_1x_2-y_1y_2,x_1y_2 + x_2y_1)$

The properties of field follow from them.

My question is: Is there a problem in defining complex numbers simply by $z = x+iy$, where $i² = -1$ and $x$, $y$ real numbers and import from $\mathbb{R} $ the operations? Or is just a elegant manner to write the same thing?

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Theoretically there COULD be a problem. Just imagine that you would define a new kind of numbers by $w=x+iy+jz$, where $i^2=1$, $ij=0$ and $j^2=-1$ (and $x,y,z$ are reals). Then you would be able to conclude that $0=\left(ij\right)^2=i^2j^2=1\cdot \left(-1\right)=-1$ in your new number system. This, of course, makes sense only if you take care to distinguish your new numbers from the old (real) ones, as the reals don't satisfy $0=-1$. This doesn't happen with the complex numbers, and this is something you must prove (even if it's easy) - you can't just pretend that it is obvious a priori. –  darij grinberg Nov 23 '11 at 1:50

4 Answers 4

up vote 16 down vote accepted

There is no "explicit" problem, but if you are going to define them as formal symbols, then you need to distinguish between the + in the symbol $a$+$bi$, the $+$ operation from $\mathbb{R}$, and the sum operation that you will be defining later until you show that they can be "confused"/identified with one another.

That is, you define $\mathbb{C}$ to be the set of all symbols of the form $a$+$bi$ with $a,b\in\mathbb{R}$. Then you define an addition $\oplus$ and a multiplication $\otimes$ by the rules

$(a$+$bi)\oplus(c$+$di) = (a+c)$ + $(c+d)i$

$(a$+$bi)\otimes(c$+$di) = (ac - bd)$ + $(ad+bc)i$

where $+$ and $-$ are the real number addition and subtraction, and + is merely a formal symbol.

Then you can show that you can identify the real number $a$ with the symbol $a$+$0i$; and that $(0$+$i)\otimes(0$+$i) = (-1)$+$0i$; etc. At that point you can start abusing notation and describing it as you do, using the same symbol for $+$, $\oplus$, and +.

So... the method you propose (which was in fact how complex numbers were used at first) is just a bit more notationally abusive, while the method of ordered pairs is much more formal, giving a precise "substance" to complex numbers as "things" (assuming you think the plane is a "thing") and not just as "formal symbols".

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There is a completely rigorous way to do the construction you allude to in the last paragraph, namely by means of quotient rings. Indeed, $\mathbb{C} \simeq \mathbb{R}[X] / (X^2 + 1)$. This generalises, for example, we can construct a commutative ring with elements of the form $x + y \epsilon$, where $\epsilon^2 = 0$. The ring so constructed is emphatically not a field, but it is sometimes useful for doing symbolic differentiation.

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Just set $i=(0,1)$ and $x=(x,0)$ for any real $x$, and the notation $x+iy$ is just a shorthand for the ordered pairs notation.

Of course you could also choose $i=(0,-1)$ ........

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Hi @Andrea. Yes, we can obtain the notation x+iy from the axioms. But it would be wrong defining first z=x+iy and import the addition and product from real numbers? I mean, there's a big mathematical fault in that idea? –  Pierre Jan 26 '11 at 18:25
    
Ok, I think that now I see better the spirit of your question. I believe that Arturo Magidin's answer fully explains the problems with your point of view. Another important approach is that given in Zhen Liu's answer. It is a useful exercise to show that the two strategies to define the complex numbers produce the same object. –  Andrea Mori Jan 26 '11 at 19:58
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@Pierre: The problem is what does "x+yi" mean? Does it denote an "expression" or simply an ordered pair, or what? See my post here for much more on this, including a jaw-dropping exchange between Cauchy and Hankel on the foundations of complex numbers. –  Bill Dubuque Jan 26 '11 at 20:13
    
@Bill Dubuque: Thanks for the reference, I'll read carefully as soon as I can. –  Pierre Jan 26 '11 at 20:55

A more geometric-flavoured construction if you're not really interested in the additive structure, but the multiplicative group $\{\mathbb{C},\times\}$.

$$\mathbb{C}\cong \mathbb{R}^+\times \mathbb{R}/2\pi\mathbb{Z},$$

via $z\mapsto (|z|,[\text{Arg}(z)])$ with the multiplication neatly defined by

$$(r,[\theta])\otimes(s,[\alpha])=(rs,[\theta+\alpha]).$$

This really simplifies De Moivre's Theorem and taking roots.

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This is off-topic. –  darij grinberg Nov 23 '11 at 1:45

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