I have a rotation of the form:
$(z(s),w(s))=B(s)(u(s),v(s))$
where $z(s),w(s),u(s),v(s)$ are in $\mathbb{R}$ and $s$ is a complex number and $B(s)$ is a $2\times 2$ matrix defined by $ B(s) = \begin{bmatrix} \cos(\theta(s)) & -\sin(\theta(s)) \\ \sin(\theta(s)) & \cos(\theta(s)) \end{bmatrix} $
The fixed point of the rotation must satisfies $(I_2-B(s))(u(s),v(s))=0$ where $I_2$ is the $2\times 2$ unit matrix. The determinant of the matrix $(I_2-B(s))$ is $-2(\cos\theta(s)-1)$ and it is not zero if $\theta (s)\ne 0\,\pmod{2\pi}$. This means that $s$ is a solution of $(u(s),v(s))=0$.
My question is what happen if $\theta(s)=0\,\pmod{2\pi}$? The reason is that this is the trivial rotation corresponding to the identity matrix, in which no rotation takes place. Does there exist zeros of $(u(s),v(s))=0$ in this case or not?