How to graph a mobius transformation, plot points of the orbit of the origin and sketch the flow line determined Problem: Let $\alpha$: $z \rightarrow \frac{(1-2i)z+1}{z+(1+2i)}$. Find the fixed points of $\alpha$. Plot the points of the orbit of the origin $\alpha^n(0), n=0,-1,-2,-3,-4,-5,1,2,3,4,5$.
Sketch the flow line determined. 
I had already shown that the fixed points of the transformation are $1$ and $-1$. I just don't know how to plot these points and what is meant by the  flow line determined. 
 A: Finding fixed points involves solving
$$z = \frac{(1-2i)z+1}{z+(1+2i)},$$
which is
$$z^2 + 4iz-1=0$$
Therefore,
$$z_{1,2} = \frac{-4i \pm \sqrt{-12}}{2} = (-2 \pm \sqrt{3})i$$
Direct calculation shows $\alpha(0) = \frac{1-2i}{5}$, next evaluation $\alpha^2(0) = \frac{-1-2i}{5}$, then $\alpha^3(0) = 0$. Negative powers are just going in the opposite direction.
A bit advanced:
since $\alpha$ has two fixed point it is conjugate to a rotation, i.e.
for $\beta$ that sends first fixed point to $0$ and second to $\infty$, $\beta\alpha\beta^{-1}$ is a rotation.
$\beta$ that sends first fixed point to $0$ and second to $\infty$ has the next form
$$\beta(z) = \frac{z + (2+\sqrt{3})i}{z + (2-\sqrt{3})i},$$
and $$\beta^{-1}(z) = \frac{(2-\sqrt{3})iz - (2+\sqrt{3})i}{-z+1}$$
Then $$\alpha^n(z) = (\beta^{-1}\circ(\beta\circ\alpha^n\circ\beta^{-1})\circ\beta)(z),$$ moreover $$\beta\circ\alpha^n\circ\beta^{-1} = (\beta\circ\alpha\circ\beta^{-1})^n$$
We then need to understand the basic rotation $\beta\circ\alpha\circ\beta^{-1}$. Let see where $1$ is send
$$\beta\circ\alpha\circ\beta^{-1}(1) = \beta(\alpha(\infty)) = \beta(1-2i) =$$
$$\frac{(1-2i)+(2+\sqrt{3})i}{(1-2i)+(2-\sqrt{3})i} = \frac{-1+\sqrt{3}i}{2},$$
therefore this is a rotation by $120^{\circ}$, which nicely explains why $\alpha^{3k} = id$.
