Is there a different way to define multiplication on Complex numbers? Let $\mathbb{C}=\{a+bi | a,b \in \mathbb{R}\}$, denote the set of complex numbers, where $i$ has the property that it's square is $-1$. 
We define the two binary operation, $+, *$ on this set $\mathbb{C}$, namely addition and multiplication of complex numbers as follows
$$(a+bi)+(c+di)=(a+c)+(b+d)i$$ and 
$$(a+bi)*(c+di)=(ac-bd)+(ad+bc)i$$
We observe that $(\mathbb{C},+,*)$ forms a mathematical object called field.
Note that $*$ has "usual" properties like distributivity, associativity and commutativity.
I have the following Question, Are there many ways of defining binary operation $*$ (or call it multiplication of complex number ), such that these above mentioned properties are satisfied. 
 A: The underlying additive group of the field $\mathbb{C}$ is just a $\mathbb{Q}$-vector space of dimension $\mathfrak{c}$.
By transport of structure, we can define a multiplication on $\mathbb{C}$ to make it isomorphic to any other such structure whose additive group is a $\mathbb{Q}$-vector space of dimension $\mathfrak{c}$.
One way to identify such is that they are precisely the $\mathbb{Q}$-algebras of cardinality $\mathfrak{c}$.
Examples of fields that we can make the result isomorphic to include:


*

*The real numbers $\mathbb{R}$

*The $p$-adic numbers $\mathbb{Q}_p$

*The field of rational Laurent series $\mathbb{Q}((x))$

*The function field $F(x)$ where $F$ is any field appearing in this list


And I've only stuck to relatively simple examples. Lots more possibilities open up if we only require the structure to be a commutive ring, or even merely a commutative $\mathbb{Q}$-algebra.

You might want to consider requiring the end result be a $\mathbb{R}$-algebra in the obvious way: that if $r$ is a real number, then $(r + 0i) * (a+bi) = (ra) + (rb)i$.
In this case, there are only three possibilities, up to isomorphism of $\mathbb{R}$-algebras.


*

*The complex numbers (e.g. $i^2 = -1$)

*The dual numbers (e.g. $i^2 = 0$)

*The hyperbolic numbers (e.g. $i^2 = 1$)


The hyperbolic numbers are isomorphic to $\mathbb{R} \times \mathbb{R}$.
Given a multiplication operation, which case you are in can be determined by looking at the minimal polynomial of $i$. Its discriminant will either be negative, zero, or positive respectively.
A: Let's not use $i$, because it is a distraction.
Given $x,y$ with $y\neq 0$, you can define:
$$(a+b\alpha)\otimes(c+d\alpha)= (ac-bd(x^2+y^2))+(ad+bc+2bdx)\alpha$$
This operation, when $x=0,y=1$, agrees with your operation.
This operation is basically gotten by thinking of $\alpha=x+yi$. Then $\alpha^2=2x\alpha -(x^2+y^2)$.
It is commutative, associative and distributive even if $y=0$. It just has zero divisors in that case. For example, if $x=1,y=0$, then $(1-\alpha)^2=0$.
When $y\neq 0$, it defines a field that is isomorphic with the complex numbers, but appears quite different.
More generally, given any real $u,v$ you can define:
$$(a+b\alpha)\otimes(c+d\alpha)=(ac+bdu)+(ad+bc+vbd)\alpha$$
These are all the cases of commutative, associative and distribute multiplications where we have that $(a+0\alpha)\otimes(c+d\alpha)=ac+ad\alpha$, so that we can think of our new ring as an extension of the real numbers.
