TITLE: Why the mapping $1/\bar{z}$ look like the field lines of an electric dipole?
QUESTION [12 upvotes]: While I was studying Mobius Transformation in  complex analysis, the complex function $\frac{1}{\bar{z}}$ caught my attention. suppose there are $n$ lines equally spaced all intersecting at $(1,0)$ as shown in red. If I apply the complex function then why does it resemble the field lines of a dipole(as shown in blue)?

I know the lines are reflected by the circle centered at $(0,0)$ and having a radius of $1$. The images of lines are circles passing through the center.
So far i have tried deriving the field line equation of a dipole but got no success. Here is my work:
Dipoles are situated at $(1,0)$ and $(0,0)$. The force acting on a unit charge at a point $(h,k)$ is equal to $\frac{1-h}{((h-1)^2+k^2)^{3/2}} + \frac{h}{((h)^2+k^2)^{3/2}} \hat{i} + \frac{-k}{((h-1)^2+k^2)^{3/2}} + \frac{k}{((h)^2+k^2)^{3/2}}\hat{j}$. I don't know how to proceed from here. My idea is that if the field equation turns out to be a that of a circle then i will be sure.
Also, why does it resemble the shape of the dipole in the first place? How can I get a feel for it. I can't see any connection between the two at all.

REPLY [2 votes]: Real electric/magnetic dipoles do not have perfectly circular field lines like here; their shape depends on the field strength. Nevertheless, since the input is a configuration of lines that is invariant under conjugation, it suffices to consider only $\frac1z$ and not $\frac1{\overline z}$.
Consider an arbitrary line passing through $(1,0)$; its polar equation is $r=\frac{\cos a}{\cos(\theta-a)}$ where $a$ is the line's normal. The $\frac1z$ map then sends
$$\left(\frac{\cos a}{\cos(\theta-a)},\theta\right)\to\left(\frac{\cos(\theta-a)}{\cos a},-\theta\right)$$
and so the equation of the transformed line is $r=\frac{\cos(\theta+a)}{\cos a}$, a circle containing the points $0$ and $1$. Varying $a$ runs through all possible circles containing $0$ and $1$, generating the observed dipole-like pattern.