I have a homework problem that is explicitly stated as follows:
The set $C$ of the complex numbers can be defined in two different ways:
a. As a set of all ordered pairs $(x,y)$ where $x,y∈R$
b. As a set of all ordered pairs $(ϕ,ρ)$ where $ϕ∈[0,2π)$ and $ρ>0$, and the pair $(0,0)$
Establish an isomorphism between these two sets, which will preserve the operation of addition and multiplication of complex numbers.
I asked this question (incorrectly) yesterday and the conclusion in the comments/answer was that the question was ill-posed. After talking to the course instructor (who was covering monoids, groups, semigroups, homomorphism, and isomorphism in that lecture), I was suggested to find a mapping $f$ such that $f$ preserved Cartesian and Polar addition, and that 'addition is harder than multiplication'. I made a new attempt and wanted to ask the community whether I made some significant progress.
I have two groups $G1$ and $G2$ of Cartesian and Polar ordered pairs respectively. Let me take the group operation for both groups to be addition, because I have to show that addition is preserved.
Then I have $f((x_1,y_1)+(x_2,y_2))=f((x_1,y_1))+f((x_2,y_2))$. What would a suitable $f$ be? The natural choice of $f$ is the mapping $f((x,y))=(arctan(\frac{y}{x}),\sqrt{x^2+y^2})$. Then, we have $$f((x_1,y_1))+f((x_2,y_2))=(arctan(\frac{y_1}{x_1}),\sqrt{x_1^2+y_1^2})+(arctan(\frac{y_2}{x_2}),\sqrt{x_2^2+y_2^2})$$ and $$f((x_1,y_1)+(x_2,y_2))=f((x_1+x_2,y_1+y_2))=(arctan(\frac{y_1+y_2}{x_1+x_2},\sqrt{(x_1+x_2)^2+(y_1+y_2)^2}))$$
While the formula for $arctan(a)+arctan(b)$ is well-known, it is easy to see that the terms under the radical signs do not agree. However, we 'know' that what we are doing is adding in polar coordinates, so it should work out 'intuitively' - but I'm not sure how to put it into words.
Any help would be appreciated!