Rotating $(3,3)$ by $45^\circ$ about center $(2,1)$ 
With a two dimensional surface, if we take $(2,1)$ as the center point and consider a transformation with a rotation axis around $45^\circ$, then point $(3,3)$ is transformed into point $(?,?)$

I am a bit stumped on how to do a $45^\circ$ rotation. I'd prefer answers that steer clear of using a rotation matrix.
 A: If we shift the origin to $(2,1)$, the coordinates of $(3,3)$ become $(1,2)$. Now multiply $1+2i$ by $e^{\frac{i\pi}{4}}$ to get the rotated coordinates to be $\dfrac{-1+3i}{\sqrt2}$. Now add $2+i$ to this and reshift the origin to $(0,0)$.
A: Let $O=(2,1)$ and $A=(3,3)$. It is easy to rotate $A$ about $O$ by 90°, 
to $A'=(0,2)$. The midpoint of $AA'$ is $M=(3/2,5/2)$ and the point $A''$ we want to find lies on ray $OM$, at a distance $OA''=OA=\sqrt2 OM$ from $O$. 
We have then:
$$
A''=O+{\sqrt2}(M-O)=(2,1)+{\sqrt2}\left(-{1\over2},\ {3\over2}\right)
=\left(2-{1\over\sqrt2},\ 1+{3\over\sqrt2}\right).
$$
A: While the rotation matrix that rotates $(x, y)$ by angle $\theta$ counterclockwise around a center $(c_x, c_y)$ may look scary,
$$\left[\begin{matrix} x^\prime \\ y^\prime \end{matrix}\right] = 
\left[\begin{matrix} c_x \\ c_y \end{matrix}\right] + \left[\begin{matrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta \\
\end{matrix}\right] \left[ \begin{matrix} x - c_x \\ y - c_y \end{matrix} \right]$$
it is only a way to write
$$\left\lbrace ~ \begin{align}
x^\prime &= c_x + (\cos\theta)(x - c_x) - (\sin\theta)(y - c_y) \\
y^\prime &= c_y + (\sin\theta)(x - c_x) + (\cos\theta)(y - c_y) \\
\end{align} \right.$$
where $(x^\prime, y^\prime)$ are the coordinates of the rotated point.

When $\theta = 45°$, $\cos\theta = \sqrt{1/2}$ and $\sin\theta = \sqrt{1/2}$.  So, to rotate point $(3, 3)$ by $45°$ around point $(2,1)$:
$$\left\lbrace ~ \begin{align}
x &= 2 + \sqrt{\frac{1}{2}}(3 - 2) - \sqrt{\frac{1}{2}}(3 - 1) \\
y &= 1 + \sqrt{\frac{1}{2}}(3 - 2) + \sqrt{\frac{1}{2}}(3 - 1) \\
\end{align} \right.$$
which is approximately $(1.293, 3.121)$.
