Show that the group of the unknotted $K=\{(z_z,z_2)\in \mathbb{S^3} : |z_1|=1 \}$ is infinite cyclic. where $\mathbb{S^3}$ is to be considering as the unit vectors in $\mathbb{C^2}\cong \mathbb{R^4}$.
My attempt was to take $A=\{(z_1,z_2)\in \mathbb{S^3}: |z_1|^{2} \leq |z_2|^{2}\}$ and $B=\{(z_1,z_2)\in \mathbb{S^3}: |z_1|^{2} \geq |z_2|^{2}\}$ and to show that they are both homeomorphic to the solid torus -which I don't know if it is easy to find a homeomorphism or not- now $A\cap B$ is a torus, and we can consider the unknotted as the center circle "center line" in $A$, so $K=\{(z_z,z_2)\in A: |z_2|=0\}$ hence the boundary of $A$ is a torus ($\cong A \cap B$ ) which is deformation retract of $A-K$, now what can I say about $B$?.
Something else which I need some help too is to understand: the torus $\cong A \cap B$ is the common boundary of both $A$ and $B$ where $A$ and $B$ are both solid tori.. I cannot imagine this! something wrong or it is hard to imagine?