TITLE: Why does the graph of $e^{1/z}$ look like a dipole?
QUESTION [43 upvotes]: I was looking at the color wheel graph of $e^{1/z}$, and my girlfriend commented that it looked just like a dipole. Does anyone have an explanation for that, why the geometry would be so similar? I guess as we follow e.g. the red color from the left side as it goes up, and then back down to the right side, we are tracing a ray outward from the origin in $\mathbb{C}$.

REPLY [16 votes]: This answer is in response to Marvis's (a.k.a. "Please delete account" a.k.a. user17762) answer, which suggests that the dipole-like behavior of $e^{1/z}$ and $e^{1/z^2}$ is due to resemblance to $1/z^2$, but looking at the color graph of $1/z^2$ itself suggests that this is not the case.
The gradient field of a dipole is $$E=\left(\frac{x^2-y^2}{(x^2+y^2)^2},\frac{2xy}{(x^2+y^2)^2}\right)=\frac1{z^2}$$ (expressing the two real variables as real and imaginary part of a single complex variable). We are interested in a function which increases in magnitude along the field lines, i.e. $f'(z)/f(z)$ is proportional to the gradient field. (Note that the original observation was that the color changes as you pass from one field line to another, and the color bands follow the field lines. The only way for an analytic function to do this is for the magnitude, but not the phase, to change as you follow the field lines.) Thus, we have the differential equation:
$$\frac{f'(z)}{f(z)}=\frac d{dx}\log f(x)=\frac c{z^2}\Rightarrow\log f(x)=a-\frac cz\Rightarrow f(x)=Ae^{-c/z},$$
and the original post had $A=-c=1$. Thus we see that it is not just any function, but precisely the function $e^{1/z}$ that looks like a dipole in the color plot. (Indeed, $e^{1/z^2}$ looks much more like a quadrupole, for the same reason.)