Plotting $0<\text{Arg}\frac{z-1}{z+1}<\frac{\pi}4$ I am supposed to draw the region in the $z$-plane for which
$$0<\text{Arg}\frac{z-1}{z+1}<\frac{\pi}4.$$
My partial attempt: First I examine the boundary where $\text{Arg}\frac{z-1}{z+1}=\frac{\pi}4$, because then the imaginary and real parts equal, so by
$$\text{Arg}\frac{z-1}{z+1}=\frac{x^2-1+y^2+2 i y}{(x+1)^2+y^2}$$
it follows that 
$$\frac{x^2-1+y^2}{(x+1)^2+y^2}=\frac{2 y}{(x+1)^2+y^2},$$
with solution
$$x^2+(y-1)^2=2.$$
This is a circle, but finding the other boundary, that is, when $\text{Arg}\cdots=0$, seems impossible, because I get $y=0$, which is clearly not possible. 
Does anybody know how to plot this region?
 A: Hint: if you know how to do conformal mapping plots for rational functions (it is the standard exercise for complex analysis courses in most universities) then you can plot the image of the cone $0<\arg(w)<\frac{\pi}{4}$ under the map $f(w)$ where
$$
w=\frac{z-1}{z+1}\quad \Leftrightarrow\quad z=\frac{1+w}{1-w}=-1-\frac{2}{w-1}=f(w).
$$
A: Notice, that when $\text{s}\in\mathbb{C}$:
$$0<\arg\left(\text{s}\right)<\frac{\pi}{4}$$
We know that $\Re\left[\text{s}\right]>0$ and $\Im\left[\text{s}\right]>0$:
$$\arg\left(\text{s}\right)=\arctan\left(\frac{\Im\left[\text{s}\right]}{\Re\left[\text{s}\right]}\right)$$

So, we get when $\text{z}\in\mathbb{C}$:


*

*$$\Re\left[\frac{\text{z}-1}{\text{z}+1}\right]=1-\frac{2\left(1+\Re\left[\text{z}\right]\right)}{\left(1+\Re\left[\text{z}\right]\right)^2+\Im^2\left[\text{z}\right]}$$

*$$\Im\left[\frac{\text{z}-1}{\text{z}+1}\right]=\frac{2\Im\left[\text{z}\right]}{\left(1+\Re\left[\text{z}\right]\right)^2+\Im^2\left[\text{z}\right]}$$


So:
$$\arg\left(\frac{\text{z}-1}{\text{z}+1}\right)=\arctan\left(\frac{\Im\left[\frac{\text{z}-1}{\text{z}+1}\right]}{\Re\left[\frac{\text{z}-1}{\text{z}+1}\right]}\right)=\arctan\left(\frac{2\Im\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]-1}\right)$$
Using, WolframALpha to plot:
$$0<\arctan\left(\frac{2\Im\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]-1}\right)<\frac{\pi}{4}$$
I got (where $\text{a}=\Re\left[\text{z}\right]$ and $\text{b}=\Im\left[\text{z}\right]$):

