Ans: for rotation and scaling factor are $\frac{\pi}{6}$ and 8 respectively.
I found a related question, but it wasn't explaining it well. I also understand scale and rotation separately but cannot figure out how to derive rotation and scaling separately. Intuitively I thought it might be a scale of 4 as it's factorable, and a 90 degree rotation of $\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}$ as the signs (+/-) were aligned(pun not intended). I was completely wrong with my intution here.
I've worked out that the eigen values and that it is diagonalizable: $A=PDP^{-1}\rightarrow A=\begin{pmatrix}-i&i\\ \:1&1\end{pmatrix}\begin{pmatrix}4\sqrt{3}+4i&0\\ 0&4\sqrt{3}-4i\end{pmatrix}\begin{pmatrix}-i&i\\ \:1&1\end{pmatrix}^{-1}$
How can we derive the rotation and scaling vector from here?
Found the below in the textbook, but cannot figure out how to apply it to find the rotation.
(My failed attempt. I did not put it in as it would lengthen the question without significant focus or contribution.)
The textbook's "Example $6$" gives you precisely the answer you need. In your case, we have $a = 4 \sqrt{3}$ and $b = -4$. We find that $r = \sqrt{a^2 + b^2} = \sqrt{48 + 16} = 8$. From what the example states, we have \begin{align} \pmatrix{4\sqrt{3} & -4\\4 & 4\sqrt{3}} &= C = r\pmatrix{a/r & -b/r\\b/r & a/r} = \pmatrix{r&0\\0&r}\pmatrix{a/r & -b/r\\b/r & a/r} \\ & = \pmatrix{8 & 0\\0 & 8} \pmatrix{\sqrt{3}/2 & -1/2\\ 1/2 & \sqrt{3}/2} = \pmatrix{8&0\\0&8} \pmatrix{\cos \varphi & -\sin\varphi\\ \sin \varphi & \cos \varphi}. \end{align} So, all that remains is to find the angle $\varphi$ such that $\cos\varphi = \sqrt{3}/2$ and $\sin \varphi = 1/2$.
Alternatively, you have already found that the eigenvalues are of the form $\lambda = 4\sqrt{3}\pm 4i$. As the example states, the scaling factor is the magnitude $|\lambda|$, and the angle is the argument of $\lambda$. In general, the argument of a complex number $a + bi$ can be found as $\arg(a + bi) = \tan^{-1}(b/a)$.