Complex numbers $z=a+ib$, what is the operation between $i$ and $b$

Question

We define a complex number as a number of the form $a+ib$, and then we can equip the set of complex numbers with addition ($+$) and multiplication ($*$), these two operations have good properties, thus $\mathbb{C}$ is a field. But it seems we did not define what is the meaning of $ib$, does it mean $i*b$ even though we define complex numbers first and then the multiplication?

In fact, we seem to always treat it as $i*b$. For example, by polar transformation $x=rcos\theta, y=rsin\theta$, we have $z=rcos\theta+irsin\theta$, but we also denote it as $r(cos\theta+isin\theta)$. Then by Euler's identity, we can get $z=re^{i\theta}$

Answer 1

This is why formal definitions are so important. As you surely can find on this forum, a more rigorous definition is to define $\mathbb{C}$ as a set to be simply $\mathbb{R^2}$, and define two operations on it:

$(a,b)+(c,d)=(a+c,b+d)$

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

These operations turn $\mathbb{C}$ into a field, which can be easily checked. Finally, we denote the pair $(a,b)$ by $a+ib$. This is just a notation, nothing else. For example, when we write $ib$, what we really mean is the pair $(0,b)$. For example, $i=(0,1)$ by definition.

Thing is, this is a very nice notation, as we get that $ib$ is indeed equal to the product of $i=(0,1)$ and $b=(b,0)$. (as you can easily prove using the definition of multiplication)

Answer 2

Yes, $ib$ is defined as $i*b$.

For $a+ib$, the order of operations applies. We multiply $i$ and $b$, then add $a$.

Similarly, in the expression $-a+b$, we treat $a$ as being multiplied by the minus sign, and then add $b$. $-a+b$ does not always equal $-(a+b)$. For example, $-1+2=1\neq-(1+2)=-3$

Answer 3

The precise meaning of $ib$ depends on the construction of the complex numbers.

If they are defined as the set of numbers of the form $a+bi$ with multiplication defined as $(a+bi)(c+di)=ac-bd+bci+adi$ then you are right that $0+bi=bi$ is not multiplication in the usual sense. However the product $(b+0i)(0+1i)=b\cdot i$ is always equal to $bi$. So we don't run into any trouble by not distinguishing between them.

There are alternative constructions of the complex numbers that avoid this ambiguity, for example we could defined numbers as pairs $(a,b)$ instead. Here $bi$ corresponds to $(0,b)$ and the multiplication operation goes $(a,b)\cdot(c,d)=(ac-bd,bc+ad)$. This is the construction lying "under the hood" in the construction you describe.

One final way to interpret $bi$ is by constructing the complex numbers as an extension of the real numbers where we introduce an extra number $i$. We require it to obey the usual properties of addition and multiplication, as well as an additional identity $i^2=-1$. Here the multiplication between $b$ and $i$ in $bi$ is the same as between a coefficient and a variable in a polynomial (with the extra identity that $i^2=-1$).