# Understanding complex number product definition

I think complex number is really just 2D vector with product defined differently. But what is the significance of the way the product is defined for complex number: $(x_1x_2-y_1y_2,x_1y_2+x_2y_1)$? Why don't we go with say this $(-y_2,x_1y_2+x_2y_1)$? We still have $i^2=-1$ right?

• Did you get your proposed product correct? Your product isn't commutative among other things. – Jonathan Feb 9 '18 at 20:12
• You should extend the real product. With your definition (1,0) x (1,0)= (0,0) – Naj Kamp Feb 9 '18 at 20:13
• It is very convenient for the set of complex numbers to satisfy the usual rules of algebra: commutativity, associtivity, distributivity, inverses, etc. Your proposed definition doesn't deliver. – Umberto P. Feb 9 '18 at 20:13
• @DanielLi other products may make $\mathbb{C}$ into a field, but they will be more complicated than the standard one (and they result in a field isomorphic to the usual complex numbers) – Jonathan Feb 9 '18 at 20:29
• @DanielLi let's put it like that: there is only one field (up to isomorphism) that you can get by taking $\mathbb R$ and adding some element $i$ such that $i^2=-1$. idok's answer basically proves that – Glougloubarbaki Feb 9 '18 at 20:31

When you deal with $(a,b)$, what you should really have in mind is $a + bi$, even if you haven't yet formally defined it that way. Now, having that in mind, the multiplication becomes completely natural from the need to satisfy distributivity, commutativity and associativity (and $i ^ 2 = -1$):

$$(a + bi)(c+di) = ac + adi + bic + bdi^2 = (ac - bd) + (ad + bc)i$$

So, in the formal definition of multiplication of complex numbers we would write

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

It is very useful to define the multiplication of the complex numbers so that you get a field. This multiplication needs to extend multiplication of the reals, be associative, have inverses for non-zero elements, and satisfy the distributive property. Defined in the usual way, the complex numbers have the incredibly useful property that every polynomial with coefficients in the complex numbers has a root. While there are other ways to define multiplication on the complex numbers (for example instead of adding $\sqrt{-1}$ to the real numbers, add a non-trivial cube root of 1) the rules for multiplying won't be nearly as nice.

We already know how to multiply sums because multiplication distributes over addition, so this is the rule $(a+b)(c+d) = a(c+d) + b(c+d) = ac+ad+bc+bd$. If it is possible (it may not be!) we would like to use such a rule.

So we don't know what $\sqrt{-1}$ is. But if it were a number as we understand them (it may not be!) it should satisfy this multiplication. Then

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

Now by our understanding of what a square root means, we want $\sqrt{-1}\cdot \sqrt{-1} = -1$, so our product is just

$$(ac-bd) + (ad+bc)\sqrt{-1}$$

A complex number is indeed just a two component vector. When we define it this way have multiplication by a complex number $a+b\sqrt{-1}$ as multiplication by the matrix

$$\pmatrix{a & -b \\ b & a}$$

If you work through the algebra of $\pmatrix{a & -b \\ b & a}\times \pmatrix{c \\ d}$ then you will see these are identical. We can also use this to demonstrate that $A\times (B + C) = A\times B + A\times C$, that $A\times(B\times C) = (A\times B)\times C$, that $A\times B = B\times A$ and that if both $a$ and $b$ are nonzero then then $A^{-1}$ exists and $A\times A^{-1} = 1$.

So this is a very fruitful definition of multiplication. It's almost forced on us by our prior understanding.