One of the properties of the rotation matrix in $\mathbb{R}^2$, $A=\begin{bmatrix}cos(\theta) & -sin(\theta)\\sin(\theta) & cos(\theta)\end{bmatrix}$ is the following:
$A^n$ corresponds to a rotation of $n \theta$
I am asked to prove this property. Is the standard induction proof below sufficient?
I ask because $n \in \mathbb{C}$ gives rotation matrices in MATLAB. Inputs such as $n=\frac{3}{4}, 3i$ seem to return rotation matrices.
Induction proof for if $A=\begin{bmatrix}cos(\theta) & -sin(\theta)\\sin(\theta) & cos(\theta)\end{bmatrix}$ then $A^n=\begin{bmatrix}cos(n\theta) & -sin(n\theta)\\sin(n\theta) & cos(n\theta)\end{bmatrix}$:
Let $n=1$ then $A^1=\begin{bmatrix}cos(\theta) & -sin(\theta)\\sin(\theta) & cos(\theta)\end{bmatrix}^1=\begin{bmatrix}cos(\theta) & -sin(\theta)\\sin(\theta) & cos(\theta)\end{bmatrix}$ which agrees.
Assume $n=k$ holds, $A^k=\begin{bmatrix}cos(\theta) & -sin(\theta)\\sin(\theta) & cos(\theta)\end{bmatrix}^k=\begin{bmatrix}cos(k\theta) & -sin(k\theta)\\sin(k\theta) & cos(k\theta)\end{bmatrix}$
For $n=k+1$, $A^{k+1}=A^kA=\begin{bmatrix}cos(k\theta) & -sin(k\theta)\\sin(k\theta) & cos(k\theta)\end{bmatrix}\begin{bmatrix}cos(\theta) & -sin(\theta)\\sin(\theta) & cos(\theta)\end{bmatrix}=\begin{bmatrix}cos(k\theta)cos(\theta)-sin(k\theta)sin(\theta) & -cos(k\theta)sin(\theta)-sin(k\theta)cos(\theta)\\sin(k\theta)cos(\theta)+cos(k\theta)sin(\theta) & cos(k\theta)cos(\theta)-sin(k\theta)sin(\theta)\end{bmatrix}=\begin{bmatrix}cos(k\theta+\theta) & -sin(k\theta+\theta)\\sin(k\theta+\theta) & cos(k\theta+\theta)\end{bmatrix}=\begin{bmatrix}cos((k+1)\theta) & -sin((k+1)\theta)\\sin((k+1)\theta) & cos((k+1)\theta)\end{bmatrix}$
which agrees.
If $n=k$ is true then $n=k+1$ is true. Since $n=1$ is true then the initial statement is true for $n \in \mathbb{Z}^+$.
If the proof above is not sufficient, how would I go about proving this property (presumably for $n \in \mathbb{C}$)?
Consider the matrix $$M(n,\theta) = \begin{bmatrix} \cos(n\theta) & -\sin(n\theta)\\ \sin(n\theta) & \cos(n\theta)\\ \end{bmatrix}, $$
for $n \in \mathbb{C}$ and $\theta \in \mathbb{R}$. We want to prove that $A^n = M(n,\theta)$.
The first thing to know to proceed is that:
$$A = \begin{bmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\\ \end{bmatrix} = e^{\theta Q}, \text{where}~ Q = \begin{bmatrix} 0 & -1\\ 1 & 0\\ \end{bmatrix}.$$
Check this link for further details on matrix exponential.
Therefore:
$$A^n = e^{n\theta Q} = M(n,\theta).$$
As a final remark, IMHO I don't feel really comfortable to define $A^n$ as a rotation matrix for $n \in \mathbb{C}$... What is the meaning of the complex angle $n\theta$?