I have to establish an isomorphism between $\Bbb R^2$ and $\Bbb C$. Specifically, I have to establish an isomorphism between the ordered pairs $(x,y)$ and $(\phi,\rho)$, where $x,y \in \Bbb R^2$ and $\phi \in [0,2\pi), \rho>0$ (including the ordered pair $(0,0)$)
I know that an isomorphism is a bijection which preserves the group operations (in this case, multiplication and addition). However, I don't quite understand how to prove it. Here is my attempt:
The ordered pair $(x,y)$ can be bijectively mapped to $x+iy$, and the ordered pair $(\phi, \rho)$ can be bijectively mapped to $$\rho \cos \phi+i \rho \sin \phi$$ Now, $(x_1+iy_1) + (x_2+iy_2)$ and $(x_1+iy_1)(x_2+iy_2)$ obviously follow the property $f(xy)=f(x)f(y)$, if my understanding of operations is correct. For addition, $f(xy)$ is $f((x_1,y_1))+f((x_2,y_2))$, and $f((x_1,y_1))=x_1+iy_1$. I can similarly do this for multiplication. Similarly, I can show the same thing for $f(\phi, \rho)$.
My question is: is this 'enough' to establish the isomorphism? I am almost sure that this proof works, but I feel like I am missing something fundamental. Any help would be appreciated!
Edit: After reading everyone's comments, I realized that I have stated the problem incorrectly. Specifically, the problem statement is as follows:
The set $\Bbb C$ of the complex numbers can be defined in two different ways:
a. As a set of all ordered pairs $(x,y)$ where $x,y \in \Bbb R$
b. As a set of all ordered pairs $(\phi, \rho)$ where $\phi \in [0,2\pi)$ and $\rho > 0$, and the pair $(0,0)$.
Establish an isomorphism between these two sets, which will preserve the operation of addition and multiplication of complex numbers.
Essentially, I have to establish an isomorphism between two different sets of ordered pairs of real numbers.
Your post is tagged group-isomorphism, so we'll go with that.
The only groups on stage are $(\mathbb R^2,+)$ and $(\mathbb C,+)$. Perhaps also the groups $(\mathbb R^{>0},\cdot)$, $(\mathbb R^\times,\cdot)$ and $(\mathbb C^\times,\cdot)$ are waiting in the wings. (The notation in the latter two means the zero element has been left out.)
There is not really a good relationship between the first two group structures and the two types of coordinates you mentioned. They simply don't play well with the operations. The normal context for a discussion of a bijjection between $\mathbb R^2$ and $\mathbb C$ is that of using the values as coordinates in the plane, so it is a more topological discussion than algebraic. You have many extraneous things to decide including restricting the domain to make it an isomorphism and figuring out what to do with $(0,0)$ and $0$.
To make any bijective correspondence compatible with algebraic operations using the map you suggested, you'd have to do something like this:
I think the best one can do with this map is to say that if $G$ the quotient group of $(\mathbb R,+)$ by the subgroup generated by $2\pi$ and $H$ is the group $(\mathbb R^{>0},\cdot)$, then $G\times H\cong (\mathbb C^\times,\cdot)$ using your map. You see, $\sin$ and $\cos$ are class functions on $G$, so you can use the elements of the quotient group as inputs for $\phi$ rather than elements of $\mathbb R$.
We know using polar decomposition that multiplication is really respected $(a,b)(c,d)=(a+c, bd)$, where the operation in the first coordinate is the quotient group addition.
IMO a more interesting bijection with the two original sets algebraically is between the $\mathbb R$ algebra $(\mathbb R^2, +,\star)$ with operations given by coordinatewise addition and multiplication $(a,b)\star(c,d)=(ac-bd, ad+bc)$ and the $\mathbb R$ algebra $(\mathbb C,+, \cdot)$, where the assignment $(a,b)\to a+bi$ is actually an isomorphism of fields.
Additional information to address edits
Let me stop you right there. Nobody defines $\mathbb C$ with item b). That is just really artificial. They do it with axioms, as a special structure on $\mathbb R^2$ (which I have already described above) or as a quotient ring $\mathbb R[x]/(x^2+1)$.
I think I've already described in detail how $(\mathbb R^2,+,\star)$ is going to work, above, and mentioned the incompatibilities with using the other map and normal operations. A major deficiency in your question here is what operations you expect to hold on these domains. In both cases, the only sensible operations are not going to be obvious ones.
If you replace the real inputs from $[0,2\pi]$ with equivalence classes, and you declare $(0,0)$ maps to $0$ by fiat, you will also have a bijection between a set and $\mathbb C$, it's just that not many people would consider the domain as a subset of $\mathbb R^2$.
And given any bijection $\phi:\mathbb C\to S$, one can just delare the operations in $S$ to be $\phi(x)+\phi(y):=\phi(x+y)$ and $\phi(x)\phi(y):=\phi(xy)$ and it is just automatically a ring isomorphism. Composing this with the inverse of the other (much more legitimate) ring isomorphism I gave earlier, you'd have a ring isomorphism between $\mathbb R^2$ and something like your other coordinate domain.