Why does the graph of $e^{1/z}$ look like a dipole? 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 $\mathbb{C}$.  
 A: Here is the color wheel chart for the exponential function from Wikipedia. Notice that the horizontal lines have "constant color",  Each of these lines can be written as $t + bi$, where $b$ is fixed and $t$ goes from $-\infty$ to $+\infty$. You notice that the constant color lines for $e^\frac{1}{z}$ are little arcs. This is just a consequence of the fact that the map $z \mapsto \frac{1}{z}$ twists the horizontal lines into the arcs. You can even see in the picture how the "left side" of the plane is mapped to the left size of 0, and the "right size" of the plane is mapped to the right side of zero. The point at $\infty$ is mapped to 0.
So you should really think of this as being a picture of what $\frac{1}{z}$ is doing.
A: Try $e^{1/z^2}$, you will get an even better match. This is due to the following:
The potential due to the dipole you have in the figure is given by $\phi(x,y) = -\dfrac{x}{x^2+y^2}$. Hence, the electric field is given as
$$\vec{E} = \left(\dfrac{x^2-y^2}{(x^2+y^2)^2}, \dfrac{2xy}{(x^2+y^2)^2}\right) \tag{$\star$}$$
If we expand $e^{1/z}-1$, we get
$$\dfrac{\bar{z}}{r^2} + \dfrac{\bar{z}^2}{2r^4} + \mathcal{O}(1/r^3)$$
If we look at the first two terms, we get
$$\dfrac1z + \dfrac1{2z^2} = \dfrac{x-iy}{x^2+y^2} + \dfrac{(x-iy)^2}{2(x^2+y^2)^2}$$
which gives us
$$\left(\dfrac{x^2-y^2+2x(x^2+y^2)}{2(x^2+y^2)^2},- \dfrac{xy+y(x^2+y^2)}{(x^2+y^2)^2}\right)$$
If you look at $e^{1/z^2}$, and truncate with the term $\dfrac1{z^2}$, you will get the exact thing as $\star$.
We have
$$e^{1/z^2}-1 = \dfrac1{z^2} + \mathcal{O}(1/z^4)$$
And
$$\dfrac1{z^2} = \dfrac{\bar{z}^2}{r^4} = \dfrac{(x^2-y^2) -2ixy}{(x^2+y^2)^2} = \left(\dfrac{x^2-y^2}{(x^2+y^2)^2}, -\dfrac{2xy}{(x^2+y^2)^2}\right)$$
which agrees with $\star$ but for the sign.
A: 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.)
