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.
Best Answer
To get circles algebraically from a dipole, we need to use an inverse law, as holds in the plane, rather than an inverse-square law. Things are pleasantly symmetric if we put the charges at the complex numbers $\pm1$. The potential is $$ V(h, k) = \log\frac{(h + 1)^{2} + k^{2}}{(h - 1)^{2} + k^{2}}, $$ whose lines of force (flow lines of the gradient) are circles through $(\pm1, 0)$ and whose equipotentials are the family of orthogonal circles.
As for the Riemann sphere pictures, here are lines through the origin and the orthogonal family of concentric circles, mapped to the sphere by stereographic projection, and then mapped back to the plane as the sphere rotates about $\pm i$:
When either $\pm1$ is carried to the north pole (the point at infinity) by rotation, the field lines become radial rays through the origin.
Here are the curves seen from directly above (with the real axis vertical), without the sphere: