Rotation and fixed points 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? 
 A: If $\theta(s) \equiv 0 \pmod{2 \pi}$, you have 
$$
B(s) = 
\begin{bmatrix}
1 & 0 \\
0 & 1 \\ 
\end{bmatrix}
$$
which is just the identity matrix. You know how to solve such a system of equations. The only point that will be mapped to $(0,0)$ by $B(s)$ is the point $(0,0)$. 
Hope that helps,
A: For each $s$ the map 
$$B(s):\quad {\mathbb R}^2\to{\mathbb R}^2,\qquad  \left[\matrix{u\cr v\cr}\right]\mapsto \left[\matrix{z\cr w\cr}\right]:=\bigl[B(s)\bigr]\left[\matrix{u\cr v\cr}\right]$$
is a counterclockwise rotation of the plane by the angle $\theta(s)$. From elementary geometry we know: If $\theta(s)\ne 0$ modulo $2\pi$ then the map $B(s)$ has a unique fixed point in ${\mathbb R}^2$, namely the origin $(0,0)$. If, on the other hand, $\theta(s)=0$ modulo $2\pi$ then $B(s)$ is the identity map of ${\mathbb R}^2$, and any point $(u,v)\in{\mathbb R}^2$ is a fixed point of $B(s)$.
In your setup the point ${\bf x}=(u,v)$ is not an independent variable, but together with $\theta(s)$ depends on the (complex) variable $s$. I understand your question as follows: "For which values of the variable $s$ is the point ${\bf x}(s)=\bigl(u(s),v(s)\bigr)$ a fixed point of the map $B(s)\>$?" On account of our preliminary considerations we now can answer this question as follows: 
The point ${\bf x}(s)$ is a fixed point of $B(s)$ iff at least one of the following two conditions is fulfilled:
$${\bf x}(s)=(0,0)\>,\qquad \theta(s)=0\ {\rm mod}\ 2\pi\ .$$
A: It seems to me that you're just confused by the terms that you introduced, so I'm going to lay out clearly what you have done.
Let $B_\theta$ be the matrix of anti clockwise rotation of $\theta$ about the origin. The matrix form of $B_\theta$ is 
$$
B_\theta = 
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta\\
\end{bmatrix}
$$
You want the fixed points of $B_\theta$, i.e. $B_\theta (u, v) = (u,v)$, which is equivalent to $(I_2 - B_\theta) (u,v) = 0$, where $I_2$ is the 2 by 2 Identity matrix. You further show that $\det(I_2 - B_\theta) = -2(\cos \theta-1) $
Now, from Linear Algebra, we know that this equation has the trivial solution $(0,0)$ if the determinant is non-zero. This corresponds to the case where $\cos \theta \neq 1$, or $\theta \neq 0 \pmod{2 \pi}$. This agrees with our understand of rotations - The only fixed point of $B_\theta$ is $(0,0)$ itself.
The 'interesting' case will be when $\det(I_2 - B_\theta) = 0$, and we can then ask what the null space of this $(I-B)$ matrix is. As mentioned, this happens when $\theta \equiv 0 \pmod{2 \pi}$. Furthermore, the null space will be the entire space, since we can actually show that $I_2 - B_{2 k \pi} = 0$. Hence, in this case, every single point will be a fixed point of $B_{2k\pi}$.
