(THIS POST HAS BEEN EDITED)
This may seem trivial, but something about this does not seem right.
So I am calculating the differential of the solenoid map:
$f: \mathbb{D} \times S^1 \to \mathbb{D} \times S^1 $ defined as $f(w,z)=(\frac{w}{8}+\frac{z}{2},z^2)$.
So, given $x=(w,z)\in \mathbb{D} \times S^1$ then $T_x(\mathbb{D} \times S^1)= T_w\mathbb{D} \times T_zS^1$.
So, what I think is, $T_w\mathbb{D}$ is a plane and $T_zS^1$ a line (i.e. you can see it generated by a vector $v_0 \in \mathbb{R}^2$).
Now, calculating the differential I get $Df_x(u,v)=(\frac{u}{8}+\frac{v}{2},2zv)$.
So what is troubling me is the second coordinate, since I am ''multiplying'' if you will, by a complex number. I think my problem is that I am not understanding too well $T_zS^1$, and probably missed some subtleties that do not let me calculate the differential in the manner I did.
Any help would be largely appreciated. Thanks in advanced.
From here I edited it:
Ok, so I am going to answer mi question:
$Df_x$ has domain the cartesian product of $T_w\mathbb{D} =\mathbb{C}$ and $T_zS^1$ which can be seen as the real vector subspace generated by $izv$, being $v \in \mathbb{R}$, then the co-domian is going to be the cartesian product of $\mathbb{C}$ and $T_f(z)S^1$, wich is actually the vector subspace generated by $iz^2v$.
So $Df_x(u,izv)=(\frac{u}{8}+\frac{iz}{2}v,2iz^2v)$. And now it makes sense, everything falls where everything needs to fall, so I think this works.
Appreciate any thoughts. T