Composition of Rotation and Translation in the Complex Plane — Finding Angle of Rotation and Point

A rotation about the point $1-4i$ is $-30$ degrees followed by a translation by the vector $5+i$. The result is a rotation about a point by some angle. Find them.

Using the formula for a rotation in the complex plane, I found $f(z)$ or the function for the rotation of the point to be $(1-4i)+(z-(1-4i))e^{-i\pi/6}$.

I know the angle of rotation for the first transformation is $-30$ or $-\pi/6$. But how do I calculate the angle of rotation for the composition? I found the angle of the vector to be $11.3$ degrees ($\arctan1/5=11.3$ degrees). Do I add $11.3$ and $-30$ to find the resulting angle of rotation? Do I also need to calculate the angle of vector 1-4i and incorporate that somehow? Is the angle of rotation for the composition simply the same as the original rotation? Would appreciate some guidance.

The rotation is defined by $$R(z)=c+e^{i\theta}(z-c)=e^{i\theta}z+(1-e^{i\theta})c,$$ and the translation by $$T(z)=z+t,$$ with $$c=1-4i,\quad \theta=-\frac\pi6,\quad t=5+i.$$ We have: \begin{eqnarray} (R\circ T)(z)&=&R(T(z))=e^{i\theta}T(z)+c(1-e^{i\theta})=e^{i\theta}z+c(1-e^{i\theta})+te^{i\theta},\\ (T\circ R)(z)&=&T(R(z))=R(z)+t=e^{i\theta}z+c(1-e^{i\theta})+t. \end{eqnarray} Let us find the fixed points of $(R\circ T)$, and $(T\circ R)$.
We have: $$(R\circ T)(z)=z\iff (1-e^{i\theta})z=c(1-e^{i\theta})+te^{i\theta}\iff z=c+t\frac{e^{i\theta}}{1-e^{i\theta}},$$ i.e. $(R\circ T)$ has one fixed point, that is \begin{eqnarray} c_1&=&c+t\frac{e^{i\theta}}{1-e^{i\theta}}=c+t\frac{e^{i\theta}(1-e^{-i\theta})}{|1-e^{i\theta}|^2}=c+t\frac{e^{i\theta}-1}{2-2\cos\theta}=1-4i+(5+i)\frac{\frac{\sqrt3}{2}-1-\frac{i}{2}}{2-\sqrt3}\\ &=&1-4i+(5+i)(2+\sqrt3)\frac{\sqrt3-2-i}{2}=1-4i-(5+i)\frac{1+(2+\sqrt3)i}{2}\\ &=&1-4i-\frac{5-(2+\sqrt3)+(10+5\sqrt3+1)i}{2}=\frac{2-(3-\sqrt3)-8i-(11+5\sqrt3)i}{2}\\ &=&\frac{-1+\sqrt3-i(19+5\sqrt3)}{2}. \end{eqnarray} Similarly $$(T\circ R)(z)=z\iff (1-e^{i\theta})z=c(1-e^{i\theta})+t \iff z=c+\frac{t}{1-e^{i\theta}},$$ i.e. $T\circ R$ has exactly one fixed point, that is: \begin{eqnarray} c_2&=&c+\frac{t}{1-e^{i\theta}}=c+\frac{t(1-e^{-i\theta})}{|1-e^{i\theta}|^2}=1-4i+\frac{(5+i)(1-\frac{\sqrt3}{2}-\frac{i}{2})}{2-\sqrt3}\\ &=&1-4i+\frac12(5+i)(2+\sqrt3)(2-\sqrt3-i)=1-4i+\frac12(5+i)[1-(2+\sqrt3)i]\\ &=&1-4i+\frac12[5+2+\sqrt3+(1-10-5\sqrt3)i]=\frac{9+\sqrt3-(13+5\sqrt3)i}{2} \end{eqnarray} Hence $R\circ T$ and $T\circ R$ are rotation with angle $-\frac\pi6$, and centers $c_1$ and $c_2$, respectively.
$(x,y)\rightarrow{\frac{1}{2}(\sqrt{3} x+y-\sqrt{3}+16),\frac{1}{2}(-x+\sqrt{3} y+4 \sqrt{3}-5)}$