Why is multiplication of complex numbers defined the way it is?

Question

I am reading in my linear algebra textbook that for two complex numbers a + bi and c + di, their product is defined to be (ac - bd) + (ad + bc)i. But this definition is stated without motivation and I do not see what that motivation is. It seems like there are many possible definitions that could’ve been chosen instead; changing the plus to a minus and/or vice versa, or switching around the order of the subtraction, or switching around which letters get multiplied by which. So what was the rationale for choosing this specific expression to represent multiplication in complex arithmetic?

If you do the calcs on $(a+bi)(c+di)$ under the assumption that multiplication is distributive over sum, commutative and associative, and that $i^2=-1$, you'll end up with that exact formula for the real and imaginary part of the product. — Sassatelli Giulio, Oct 17 '23 at 23:17
The negative sign is there because $(bi)(di)=-bd$

assuming $i^2=-1$ — WW1, Oct 17 '23 at 23:18
@WW1 moreover, if you replace -1 with something else (such as 0 or +1), you'll get a completely different algebra. — Trang Oul, Oct 18 '23 at 13:24

score 23 · Answer 1 · answered Oct 17 '23 at 23:22

What linear algebra textbook is this and where in the book is multiplication with complex numbers introduced?

Essentially all that is going on in what you asked about is that we want multiplication to distribute over addition and we want $i^2 = -1$. There really is nothing more to it than that.

Here are details. When we have two linear polynomials $a+bx$ and $c+dx$, here is their product: $$ (a+bx)(c+dx) = ac + a(dx) + (bx)c + (bx)(dx) = ac + (ad + bc)x + bdx^2. $$ Now suppose we want to multiply $a+bi$ and $c+di$ in a way that makes the usual rules involving addition and multiplication work but also require $i^2 = -1$. Then $$ (a+bi)(c+di) = ac + a(di) + (bi)c + (bi)(di) = ac + (ad + bc)i + bdi^2. $$ Since we want $i^2 = -1$, that last term $bdi^2$ is $-bd$, so $$ (a+bi)(c+di) = ac + (ad + bc)i - bd = (ac - bd) + (ad+bc)i. $$ That's it.

score 1 · Answer 2 · answered Oct 18 '23 at 20:51

It seems natural to ask that $\mathbb C$ is a field with the following properties:

$\mathbb R$ is a subfield of $\mathbb C$.
There is an $i\in\mathbb C$ such that $i^2=-1$ and for all $z\in\mathbb C$, there exists a unique pair of real numbers $a,b$ such that $z=a+bi$.

These two properties leave us with no choice for how addition and multiplication should be defined: it must be the case that $(a+bi)+(c+di)=(a+c)+(b+d)i$ and that $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$.

score 1 · Answer 3 · answered Oct 19 '23 at 17:02

The main defining feature of complex numbers is that $i^2=-1$. If we assume that addition and multiplication work the same way for real and complex numbers, we can derive the identity from the textbook.

If we start with $(a+bi)(c+di)$, we can use algebra to simplify to $ac+adi+bci+bdi^2$. Substituting $i^2$ for $-1$ gives us the form $ac+adi+bci+bd(-1)$. This is a perfectly fine identity as is, but the book rearranged it a little for clarity.

$ac+adi+bci+bd(-1)$ $=ac+adi+bci-bd$ $=(ac-bd)+(adi+bci)$ $=(ac-bd)+(ad+bc)i$

The specific form is arbitrary, it's entirely up to the textbook author and what they think will be easiest for readers to understand.

Why is multiplication of complex numbers defined the way it is?

3 Answers3