Complex multiplication definition

Question

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

I'm in a book dealing with quaternions and it says the above is the definition of multiplication for complex numbers.

Can someone show me the proof?

Answer 1

Technically this is a definition, but here's the motivation. Complex numbers are thought of as numbers in the form of $a+bi$, where $a$ and $b$ are real numbers, and $i$ has the property that $i^2=-1$. Then using the distributive law, $$(a+bi)\cdot (c+di) = a\cdot c+a\cdot di + bi\cdot c + bi\cdot di =$$ $$ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i$$ Representing $a+bi$ simply as $(a,b)$, this translates into your above formla.

Answer 2

This looks more like complex numbers than quaternions (which should have 4 components each), although it works if the second component is any one of $i, j, k$.

If we swap the notation, it becomes a bit more obvious:

$\begin{eqnarray}(a, b)(c, d) & = & (a + ib)(c + id) \\ & = & a(c + id) + ib(c + id) \\ & = & ac + iad + ibc + i^2 bd \\ & = & ac + iad + ibc - bd \mbox{ since } i^2 = -1 \\ & = & ac - bd + iad + ibc \\ & = & (ac - bd) + i(ad + bc) \\ & = & (ac - bd, ad + bc)\end{eqnarray}$

Answer 3

What you expect for real $a,b,c,d$ is that complex multiplication gives

$$(a+b\sqrt{-1})(c+d\sqrt{-1}) = ac +ad \sqrt{-1}+ bc\sqrt{-1} +bd\sqrt{-1}^2 = (ab-bd) +(ad+bc)\sqrt{-1}$$

but if you want to avoid using an as-yet-undefined $\sqrt{-1}$, then defining multiplication on the ordered pairs of reals in this way produces the desired result in terms of previously defined terms and you can write $(0,1)$ instead of $\sqrt{-1}$

Complex addition is then the simpler $(a,b)+(c,d)=(a+c,b+d)$