# 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?
• As for your first question: A complex number can be regarded as a vector with the real part/component along the "$x$-axis" and imaginary part along the "$y$-axis", so the $i$ is just a way of distinguishing between the two dimensions of the complex number/vector; you can instead distinguish these by using the unit vectors of $x$ and $y$. – Bobson Dugnutt Jun 16 '17 at 10:26
• @Lovsovs Thank you for your comment. Perhaps I am just confused here and $e^\theta = sin(\theta) + cos(\theta)$ also applies without i included? I'm heaving hard time to get my head around why Real numbers can be rotated from a formula defined in the the domain of Complex numbers. – 01000001 Jun 16 '17 at 10:39
• Use & to separate matrix elements that are on the same row so that they donâ€™t run together. – amd Jun 17 '17 at 0:19

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
$$\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} ac-bd & -(ad+bc) \\ ad+bc & ac-bd \end{pmatrix}$$
• Thank you for your answer. I seem to understand why this matrix representation work, I just can't get my head around why $i$ can be emitted, or more precisely why can we apply a rotation matrix - derived from a formula with complex numbers - on real numbers. – 01000001 Jun 16 '17 at 10:42