The a + ib Form: Real Part, Imaginary Part, and the Symbol i

Ordered pairs are precise but clumsy to write, so the next step is to recover the familiar $a+ib$ notation from them — and to show it means exactly the same thing.

Notice that for any real numbers $a, b$: $(a,0) + (b,0) = (a+b, 0)$ and $(a,0)\cdot(b,0) = (ab, 0)$. In other words, complex numbers whose second entry is $0$ behave under addition and multiplication exactly like ordinary real numbers. This lets us simply identify the real number $a$ with the complex number $(a,0)$, writing $a$ in place of $(a,0)$ without any loss of information. Every real number is now also a complex number.

Next, give the special complex number $(0,1)$ its own name: $i$. Compute its square using the multiplication rule from Section 1.1:

So $i^2 = -1$ — the multiplication rule we defined in Section 1.1 was engineered precisely so that this would hold. This is the 'missing' square root of $-1$ that motivated the whole chapter, now sitting on solid logical ground.

Any complex number $(a,b)$ can now be rewritten: $(a,b) = (a,0) + (0,b) = (a,0) + (b,0)\cdot(0,1) = a + ib$. So instead of writing $z=(a,b)$, we write $z = a+ib$, where $a$ is called the real part of $z$ (written $\operatorname{Re}(z)$) and $b$ is the imaginary part (written $\operatorname{Im}(z)$) — note that the imaginary part is itself a real number, it's just the coefficient sitting next to $i$. A complex number with $\operatorname{Im}(z)=0$ is a real number; one with $\operatorname{Re}(z)=0$ (and $b\ne0$) is called purely imaginary.

Under this identification, the addition/subtraction/multiplication rules from Section 1.1 turn into the ordinary algebra you'd expect if you just treat $i$ as a quantity satisfying $i^2=-1$ and expand normally:

That second line is worth pausing on: you never need to memorise the product rule $(ac-bd, ad+bc)$ separately — just multiply out $(a+ib)(c+id)$ like two binomials and replace $i^2$ by $-1$ wherever it appears.

Division needs a trick, because there's no direct 'divide entry-wise' rule. To simplify $\dfrac{1}{c+id}$, multiply top and bottom by $c - id$ (the sign of the imaginary part flipped) so the denominator becomes real:

$\frac{1}{c+id} = \frac{c-id}{(c+id)(c-id)} = \frac{c-id}{c^2+d^2}$ …