Given matrix A. To find $A^{2010}$ 
Let $\theta = 2\pi/67$. Now consider the matrix
  $$
 A =
 \begin{pmatrix}
  \phantom{-}\cos \theta & \sin \theta \\
  -\sin \theta & \cos \theta
 \end{pmatrix}.
$$
  Then the matrix $A^{2010}$ is
\begin{align*}
 &\text{(A)}\;
 \begin{pmatrix}
  \phantom{-}\cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta
 \end{pmatrix},
 &
 &\text{(B)}\;
 \begin{pmatrix}
  1 & 0 \\
  0 & 1
 \end{pmatrix}, \\
 &\text{(C)}\;
 \begin{pmatrix}
  \phantom{-}\cos^{30} \theta & \sin^{30} \theta \\
   -\sin^{30} \theta & \cos^{30} \theta
 \end{pmatrix},
 &
 &\text{(D)}\;
 \begin{pmatrix}
  \phantom{-}0 & 1 \\
  -1 & 0
 \end{pmatrix}.
\end{align*}

I think answer is B. But I am not sure.
 A: Note that $A$ is a rotation matrix that takes a vector and rotates it by $1/67$th of a revolution. Now notice that $2010 = 67 \times 30$, so the answer is indeed (B).
A: Note that
$$
A=PDP^{-1}
$$
where
\begin{align*}
P &=
\begin{bmatrix}
i & -i \\
1 & 1
\end{bmatrix}
&
D &=
\begin{bmatrix}
\cos\theta+i\sin\theta & 0 \\
0 & \cos\theta-i\sin\theta
\end{bmatrix}
\end{align*}
De Moivre's formula then implies
\begin{align*}
A^{2010}
&= PD^{2010} P^{-1} \\
&= P
\begin{bmatrix}
\cos(2010\,\theta)+i\sin(2010\,\theta) & 0 \\
0 & \cos(2010\,\theta)-i\sin(2010\,\theta)
\end{bmatrix}P^{-1}\\
&=
P
\begin{bmatrix}
\cos(60\,\pi) & \sin(60\,\pi) \\
-\sin(60\,\pi) & \cos(60\,\pi)
\end{bmatrix}
P^{-1} \\
&=
P
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
P^{-1} \\
&= PIP^{-1} \\
&= PP^{-1} \\
&= I
\end{align*}
A: $A$ can be thought as $e^{-i\theta}$. So that for the natural homeomorphism $\phi:\mathbb{C}\rightarrow \mathbb{R}^2:x+iy \mapsto (x,y)$, $(\phi^{-1}\circ A\circ \phi )(z)=e^{-i\theta } z$. Hence, $(\phi^{-1} \circ A^{2010} \circ \phi)(z)= e^{- 2010 i \theta} z = e^{-60\pi i}z=z$. Hence, the answer is (B).
