Plotting complex arcs on an Argand diagram What would $arg(\frac{z-4i}{z+2i}) = \frac{\pi}{4}$ look like on an Argand diagram and why?
I don't know how to check my answer because Desmos does not support complex numbers. 

We have a Möbius transformation 
$$z'=\frac{z-4i}{z+2i}$$
which maps circles to circles or straight lines, and straight lines to circles or straight lines. We are looking for the shape of the set of complex numbers which will be mapped to a straight line of slope $1$. We suspect that the shape is that of a circle.
Let $z=a+ib.$ Then
$$\frac{a+ib-4i}{a+ib+2i}=\frac{(a+ib-4i)(a-ib-2i)}{(a+i(b+2))(a-i(b+2))}=$$
$$=\frac{a^2+b^2-2b-8-i6a}{a^2+(b+2)^2}.$$
The argument is $\frac{\pi}4$. So,
$$\operatorname {atan}\left(-\frac{6a}{a^2+b^2-2b-8}\right)=\frac{\pi}4.$$
Taking the tangent of both sides we get
$$-\frac{6a}{a^2+b^2-2b-8}=1.$$
Thus
$$a^2+b^2-2b+6a-8=(a+3)^2+(b-1)^2=18.$$
This is a circle centered at $-3+i$ and of radius $\sqrt{18}.$
EDIT
Here is the plot of the locus of the $z$'s sought:

So, negative $a$'s are considered.
