Give a $2D$ extension field of $\Bbb R_{\gt0},$ called $\Bbb B,$ s.t.
$$ \Bbb B \cong \Bbb C. $$
and $\Bbb B$ is identified with the real vector space $\Bbb R_{\gt 0} \times \Bbb R_{\gt 0}$ as a subspace.
Describe rotation in $\Bbb B.$ (I can already explicitly describe rotation in $\Bbb R_{\gt 0}\times \Bbb R_{\gt 0}$ It is based on the parametrization $(e^{\cos(\theta)},e^{\sin(\theta)}$).
Here's a diagram showing the real plane and the positive real plane and the positive real plane sitting in the real plane:
See: Explicit isomorphism? for more on why it's not-obvious how to explicitly give the isomorphism.
See: Proof that $\Bbb R_{\gt 0}$ is a vector space, for proof that $\Bbb R_{\gt 0}$ is a vector space.
So the goal is to do a $2-$dimensional field extension of $\Bbb R_{\gt 0} $ which is isomorphic to the complex plane.
See: Proof that $\Bbb R_{\gt 0}$ is a field.
As you can see I've spent a lot of time setting up this problem, so that I can ask the above question.
Andrew D. Hwang makes it clear in their answer and comments for the first link, that it's not at all obvious how to give the explicit isomorphism. Is it impossible? Is the best one can say is that they are isomorphic?
I may not understand your entire question, but I can answer the part at the top. Define $\mathbb B=\mathbb R_{>0}\times \mathbb R_{>0}.$ The isomorphism $\phi:\mathbb B\to \mathbb C$ is given by $\phi(x,y)=\ln x+i\ln y.$ Suppose we want to rotate $(x,y)$ by $\theta$ degrees counterclockwise. In $\mathbb C$ the point $(x,y)$ corresponds to $\ln x + i\ln y.$ Rotating by $\theta$ degrees gives $$ (\ln x +i \ln y)(\cos \theta + i \sin \theta)=(\ln x \cos \theta - \ln y \sin \theta) +i (\ln y\cos \theta + \ln x \sin \theta). $$ Mapping this point back to $\mathbb B$ gives $$ (e^{\ln x \cos \theta - \ln y \sin \theta}, e^{\ln y \cos \theta + \ln x \sin \theta})=(x^{\cos \theta}/y^{\sin \theta},x^{\sin \theta}y^{\cos \theta}). $$