Page-199 of VDG
I'm trying to figure out how to get the dipole flow equation as shown in the above picture from the complex function $f(z)=z^2$. I write it in components and get:
$$ u= x^2 - y^2$$
$$ v= 2xy$$
Now, how do I use this to find the integral curves?
The integral curves of the system
$\dot x= u=x^2-y^2$ and $\dot y = v=2 xy$
can be found by first expressing the system in consolidated form as a single condition on complex quantities: the complex velocity vector $ \dot z= \dot x+ i \dot y $ must be identicalto $ u+ iv=f(z)=z^2$ at each instant, hence $ \dot z= \lambda(t) z^2$ where $\lambda(t)=1$ at each time.
Separating variables and integrating, deduce that $ \int \frac{ dz/dt }{z^2} \ dt= - z^{-1}(t) + z^{-1}(t_0) = t- t_0$ This can be solved for $z(t)$ to obtain a parametrization of each integral curve in terms of its initial position $z_0$.
More succinctly, the imaginary part of $- z^{-1}(t) + z^{-1}(t_0) $ must vanish, and this gives an implicit equation for the family of all such integral curves. Looking at the diagram it looks like you might choose $z_0$ to lie always on the imaginary axis to generate a one-parameter family of such integral curves: $ \Im( z^{-1})= C$. That is, $y= C (x^2+y^2)$ which describes a family of circles centered on the $y$ axis, all passing through the origin.