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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?

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    $\begingroup$ Did you get your proposed product correct? Your product isn't commutative among other things. $\endgroup$ – Jonathan Feb 9 '18 at 20:12
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    $\begingroup$ You should extend the real product. With your definition (1,0) x (1,0)= (0,0) $\endgroup$ – Naj Kamp Feb 9 '18 at 20:13
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    $\begingroup$ 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. $\endgroup$ – Umberto P. Feb 9 '18 at 20:13
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    $\begingroup$ @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) $\endgroup$ – Jonathan Feb 9 '18 at 20:29
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    $\begingroup$ @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 $\endgroup$ – Glougloubarbaki Feb 9 '18 at 20:31
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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

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

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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.

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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.

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