I have an ambiguity about how to orient the boundary of a manifold.
In particular : Consider the example $M=B^2 \subset \mathbb{R^2}$ be the manifold with boundary. suppose positive orientation for M is induced by the positive orientation of $\mathbb{R^2}$ ({$(1,0),(0,1)$} is positive). According to Guillemin & Pollack text we must orient $\partial B^2=S^1$ such that : If $\mathcal{B}$ is a basis for $T_x(S^1)$, $sign \mathcal{B} = sign \{ n_x , \mathcal B \}$ as a basis for $T_x(B^2)$. where $n_x$ is the outward unit normal vector.
Now my question : let $x=e^{i\pi/4} \in S^1$. Now, $n_x=e^{i\pi/4}$ (based at $x$). I think either of the vectors $e^{i3\pi/4}$ and $e^{-i\pi/4}$ (both based at x) extend to positively oriented bases for $T_x S^1$, denote them by $\mathcal{B_1}$ and $\mathcal{B_2}$ respectively. Rotation by $\pi/4$ of {$(1,0),(0,1)$} gives us $\{ n_x , \mathcal B_1 \}$ and Rotation by $-\pi/4$ of {$(1,0),(0,1)$} gives us $\{ n_x , \mathcal B_2 \}$ . And we know that Rotation matrices have positive determinants $(det R_{\theta}=1)$ which implies that {$(1,0),(0,1)$} and $\{ n_x , \mathcal B_1 \}$ and $\{ n_x , \mathcal B_2 \}$ are all equivalently oriented.
I would be very thankful if you can clear up this ambiguity for me. Thanks!