Rotation matrices is basicaly what you want.

For rotating a vector
$\begin{pmatrix} x \\ y \end{pmatrix} $
by the angle $\alpha$, just multiply it with the matrix
$\begin{pmatrix} \cos(\alpha) &-\sin(\alpha) \\ \sin(\alpha) &\cos(\alpha) \end{pmatrix}$.

\begin{align}
\begin{pmatrix} x_{new} \\ y_{new} \end{pmatrix} &= \begin{pmatrix} \cos(\alpha) &-\sin(\alpha) \\ \sin(\alpha) &\cos(\alpha) \end{pmatrix} \cdot \begin{pmatrix} x_{old} \\ y_{old} \end{pmatrix} \\
&= \begin{pmatrix} x_{old} \cdot \cos(\alpha) - y_{old} \cdot \sin(\alpha) \\ x_{old} \cdot \sin(\alpha) + y_{old} \cdot \cos(\alpha) \end{pmatrix}
\end{align}

In case you are not familiar with vectors, this means:

\begin{align}
x_{new} &= x_{old} \cdot \cos(\alpha) - y_{old} \cdot \sin(\alpha) \\
y_{new} &= x_{old} \cdot \sin(\alpha) + y_{old} \cdot \cos(\alpha)
\end{align}

A vector is an element of a vector space. An element of a vector space can be an $n$-tuple of numbers, a polynomial, a matrix, a function etc.

A linear transformation transforms a vector ($n$-tuple, polynomial, matrix, function, etc.) into another vector ($n$-tuple, polynomial, matrix, function, etc.). A matrix cannot transform a vector into another vector, because you can multiply a matrix by an $n$-tuple, but you can't multiply a matrix by a polynomial, a matrix (well, not always, see below), a function, etc.

A matrix associated to a linear transformation can only multiply $n$-tuples of coordinates respect to a basis, and the results are $n$-tuples of coordinates respect to a basis.

Imagine that your vector space is the set of all symmetric $2\times 2$ matrices, and that your linear transformation is:$$T\left(\begin{bmatrix} a & b \\ b & c \end{bmatrix}\right)=\begin{bmatrix} c & a \\ a & b \end{bmatrix}$$

The simplest basis is: $\left\{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right\}$.

Respect to this basis the coordinates of $\begin{bmatrix} a & b \\ b & c \end{bmatrix}$ are $(a,b,c)$, the coordinates of $\begin{bmatrix} c & a \\ a & b \end{bmatrix}$ are $(c,a,b)$.

The matrix associated to $T$ respect to that basis is: $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$.

You can't multiply $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ by $\begin{bmatrix} a & b \\ b & c \end{bmatrix}$, but:
$$\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix}=\begin{bmatrix} c \\ a \\ b \end{bmatrix}$$
i.e.
$$\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\text{Coord}\left(\begin{bmatrix} a & b \\ b & c \end{bmatrix}\right)=\text{Coord}\left(\begin{bmatrix} c & a \\ a & b \end{bmatrix}\right)$$
This is why:

- you always need a basis to associate a matrix to a linear
transformation (when the basis is omitted you assume the canonical basis),
- the matrix associated to a linear transformation is unique respect to
a fixed basis,
- you can also have different bases for the domain and the range of a
linear transformation, so the matrix associated to a linear transformation is unique respect to the basis of its domain and the basis of its range,
- since there are infinite bases, there also are infinite matrices
associated to a linear transformation.

## Best Answer

A circle with radius $r$ in the $\mathbb R^2$ is $\{(x,y)\in\mathbb R^2:x^2+y^2=r^2\}$.

This is no different in the complex world; that is, a circle with radius $r$ in $\mathbb C$ is $\{x+iy\in\mathbb C:x^2+y^2=r^2\}$.