I have been thinking about viewing the geometric intuition of multiplying two vectors as applying linear transformations to one of the operands. Apropos of multiplying by the imaginary unit $i$ to perform a CCW rotation of $\frac{\pi}{2}$ radians on a vector.
Let's say I have two complex numbers $v_1$ and $v_2$ written in polar form below.
$$ \begin{aligned} v_1 &= 3\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) &= \begin{bmatrix} \frac{3\sqrt{3}}{2}\\ 3/2 \end{bmatrix}\\ v_2 &= 4\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) &= \begin{bmatrix} 2\\ 2\sqrt{2} \end{bmatrix}\\ \end{aligned} $$
One can obtain their product by multiplying their moduli and summing their arguments.
$$ v_1v_2 = 12\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = \begin{bmatrix} 0\\12 \end{bmatrix} $$
Shouldn't this be the same as performing two linear transformations on $v_2$? A rotation by $\frac{\pi}{6}$ radians and a scaling by 3? (I don't think the order should matter for a rotation and scaling.) When I do this, I do not get the same result as above.
$$ \begin{aligned} & \begin{bmatrix} 3\cos\frac{\pi}{6} & -3\sin\frac{\pi}{6}\\ 3\sin\frac{\pi}{6} & 3\cos\frac{\pi}{6}\\ \end{bmatrix} \begin{bmatrix} 2\\ 2\sqrt{2} \end{bmatrix}\\ =&\,2\begin{bmatrix} 3\cos\frac{\pi}{6}\\ 3\sin\frac{\pi}{6} \end{bmatrix} + 2\sqrt{3}\begin{bmatrix} -3\sin\frac{\pi}{6}\\ 3\cos\frac{\pi}{6}\\ \end{bmatrix}\\ =&\,\begin{bmatrix} 6\cos\frac{\pi}{6} - 6\sqrt{3}\sin\frac{\pi}{6}\\ 6\sin\frac{\pi}{6} + 6\sqrt{3}\cos\frac{\pi}{6}\\ \end{bmatrix}\\ =&\,\begin{bmatrix}0\\12\end{bmatrix} \end{aligned} $$
Something is wrong with my intuition here. I would expect these to be equivalent.
EDIT: Originally I had $\sin\frac{\pi}{3} = 2\sqrt{2}$. It should be $2\sqrt{3}$. Corrected. This solves the problem.