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