Answered

2022-01-17

Is there a problem in defining a complex number by

$z=x+iy?$

Answer & Explanation

nick1337

Expert

2022-01-19Added 573 answers

Step 1
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 $a,\text{}b\in \mathbb{R}$ . Then you define an addition $\oplus $ and a multiplication $\otimes $ by the rule
$(a+bi)\oplus (c+di)=(a+c)+(c+d)i$
$(a+bi)\otimes (c+di)=(ac-bd)+(ad+bc)i$
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 $+,\text{}\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.

Vasquez

Expert

2022-01-19Added 457 answers

Step 1
There is a completely rigorous way to do the construction you allude to in the last paragraph, namely by means of quotient rings. Indeed,
$\mathbb{C}\cong \frac{\mathbb{R}[X]}{{X}^{2}+1}$
This generalises, for example, we can construct a commutative ring with elements of the form
$x+y\u03f5$
where
${\u03f5}^{2}=0$
The ring so constructed is emphatically not a field, but it is sometimes useful for doing symbolic differentiation.

alenahelenash

Expert

2022-01-24Added 366 answers

Just set $i=(0,\text{}1)$
and
$x=(x,\text{}0)$
for any real x, and the notation
$x+iy$
is just a shorthand for the ordered pairs notation. Of course you could also choose $i=(0,\text{}-1)$

Most Popular Questions