Coplanar conducting sheets at different potentials There is an exercise in Zangwill (7.23 Contact Potential) shown below, it is not very clear in the diagram, but $\rho$ is normal to $z$. The aim is to compute the potential in space, you are suggested to argue that the potential is univariate in $\phi$ by symmetry arguments.
Consider a cylindrical coordinate system with axis along the boundary between the sheets. The radial direction away from the boundary has no scale, so we expect no variation in that direction. There is also symmetry along the line of the boundary, so we expect no dependence there either. This leaves just the angular variable.

I have two results for the potential of this system, but they aren't the same. This violates uniqueness.
The textbook solution is that the problem is univariate in $\phi$ and therefore linear in $\phi$. And if it's linear in $\phi$ then the $E$ field lines have power law spacing.
However, when I did this, I thought of a solution via conformal map where I take the solution of a infinite parallel plate capacitor and map the coordinates with a function so that the two plates become co-linear like in the problem statement. (Credit on Desmos tool found here made by Youtube: Partial Science, I had to make very few edits but it was useful for visualizing with a parameter to map continuously)
Starting with the infinite parallel plate capacitor, each sheet on $z=x\pm i\pi/2$, shown below,

the same curve in the mapped domain, $w=\exp(z)$ we get the upper/lower plate on $w=\pm iv, v=\exp(x)$ where the gap near zero vanishes as the lower limit of x approaches $-\infty$

But under this map, the field lines start at uniform spacing then get exponential spacing disagreeing with the first solution. I looked up the exponential map and found that it is conformal. I've run across some other properties on a region of $\mathbb C$ near zero, but none of them make it clear to me that this should fail.
What am I missing?
Turns out I messed up the method. I don't know if I should or shouldn't answer my own question if I just made a mistake.
 A: I had some misunderstanding of the conformal map technique with the complex potential. So I'll be going through the solution, my misunderstanding, and clarifying the technique.
Before this, a known solution to the problem is that $\varphi$ is a linear function of the angle around the edge of contact, $\varphi(\theta)=V\theta/\pi$ and uniqueness implies any other solution must be equal to this one. Note that the $E$ field for such a solution would be inverse to the distance from the edge of contact.
The solution to the problem (the potential as a function of space for two half planes in contact) via using conformal maps is as follows.
Consider an infinite parallel plate capacitor with distance $\pi$ between the surfaces. One surface at $\varphi=V$ the other at $\varphi=0$. Each surface is a contour described by $z = x\pm i\pi/2$. If we want these two curves to lie parallel to one another and intersect at one point, then we can use the map,
$$w=\exp(z)=\exp(x)e^{\pm i\pi}=\pm i\exp(x)$$
and these two contours will both lie in the purely complex axis and share one point. It occurs as $x\rightarrow-\infty$ where $w\rightarrow0$.
My misunderstanding happened here. I have one solution that has $\exp(x)$ and I considered this to be related to the "spacing" of field lines and therefore strength of the E field, but from the prior result, I expected field strength to be inverse to length, not exponential.
Turns out, I jumped the gun on a conclusion. The conformal map of the coordinates is not a complete description of the problem. The technique is to leverage a change of coordinates and the following fact,
$$
f(z) = f(z(w)) = \varphi + i\psi
$$
for a potential, $\varphi$, and a function, $\psi$ of field lines to describe the "same potential" in different coordinates.
Where in both cases the physical potential, $\varphi$ is described in both representations of $f$, but each in a different coordinate system. So I describe $\varphi$ in the $z$ coordinate system.
The function, $f(z)=-(V/\pi)iz$ has uniform vertical field lines, just like a parallel plate capacitor with horizontal plates. But $z=\log w$, so
$$
f(z) = -(V/\pi)iz = f(z(w)) = -(V/\pi)i \log w = f(w)
$$
and if we polar parametrize $w=r\exp(i\theta)$, then
$$
f(w) = (V/\pi) (\theta-i\log r)
$$
which has a real part $\varphi = {\mathrm{Re}}\,f =(V/\pi) \theta\,$ as expected. The imaginary part gives that there are logarithmic spacing of field lines, which I did expect, but misinterpreted.

So my misunderstanding was twofold, the first part was I didn't get to the end of the technique to find a potential, the other part was that the spacing of the field lines (related to the exponential) was not the form of the field strength (inverse power).
