<s>[INST] How to derive the Rotation Matrix from the Euler Formula I'm trying to understand how the two dimensional rotation matrix (i.e. $R \in \mathbb{R}^2$) can be derived from the Euler Formula ($e^{i\theta} = \cos \theta + i \sin \theta$). $R$ is given as:
$$
R(\theta) =
\begin{bmatrix}
  \cos\theta & -\sin\theta\\
  \sin\theta & \cos\theta
 \end{bmatrix}
$$
$$
\begin{bmatrix}
  x' \\
  y'
 \end{bmatrix} =
\begin{bmatrix}
  \cos\theta & -\sin\theta\\
  \sin\theta & \cos\theta
 \end{bmatrix}
\begin{bmatrix}
  x \\
  y
 \end{bmatrix}
$$
$$
  x' = x \cos \theta - y \sin \theta 
$$
$$
  y' = x \sin \theta + y \cos \theta
$$
My questions are:


*

*Why can be $i$ omitted from the rotation matrix? (I tried to look for explanations 1,2 but none of these explanations goes beyond that  $i$ is omitted)

*Why can we derive a rotation matrix for $\mathbb{R}^2$ from a form that is defined in $\mathbb{C}^2$? How comes we don't get complex numbers as a result after some rotations?

 [/INST] The complex number $a+bi$ can be represented by the matrix $\displaystyle \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$.
Note that $(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i$ and  
$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \pm\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} a\pm c & -(b\pm d) \\ b\pm d & a\pm c \end{pmatrix}$$
Also we have $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ and 
$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix}\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} ac-bd & -(ad+bc) \\ ad+bc & ac-bd \end{pmatrix}$$
</s>