I am trying to figure out how two expressions for the potential of a line charge between two grounded parallel conductors are equivalent. Showing the equivalence of the two expressions seems more mathematical than physical in nature, so I thought I'd post it here. The setup for the problem is that an infinite line charge $\lambda$ in the z direction at $(0,d)$ is situated between two infinite parallel grounded conducting planes, one at $y=0$ and the other at $y=a$.
The potential can be found two ways, one using Laplace's equation to obtain a series solution and the other using the method of images. I can show how these two expressions for the potential were obtained if needed, but I don't think it's relevant to my question, which is really more concerned with math than physics. The potential given by Laplace's equation is: $$\Phi(x,y)=\frac{\lambda}{\pi \varepsilon_0}\sum_{n=1}^{\infty}\frac{1}{n}\sin(\frac{n\pi d}{a})\sin(\frac{n\pi y}{a})e^{\frac{-n\pi |x|}{a}} $$ The potential given by the method of images is: $$\Phi(x,y)=\frac{\lambda}{4\pi \varepsilon_0}\sum_{n=-\infty}^{\infty}\ln(\frac{x^2+(y-2na+d)^2}{x^2+(y-2na-d)^2})$$
Using $\sin(u)\sin(v)=\frac{cos(u-v)-cos(u+v)}{2}$, the Laplace solution can be written as the real part of two summations of a complex variable: $$\Phi(x,y)=\frac{\lambda}{2\pi \varepsilon_0}\Re\{\sum_{n=1}^{\infty}\frac{Z_1^n}{n}\}-\frac{\lambda}{2\pi \varepsilon_0}\Re\{\sum_{n=1}^{\infty}\frac{Z_2^n}{n}\}$$ where $Z_1=e^{\frac{\pi}{a}(-|x|+i(d-y))}$ and $Z_2=e^{\frac{\pi}{a}(-|x|+i(d+y))}$. Using the identity that $\sum_{n=1}^{\infty}\frac{Z^n}{n}=-\ln(1-Z)$ and $\Re\ln(1-z)=\ln(|1-z|)$, I was able to obtain an expression for the potential without an infinite summation:
$$\Phi(x,y)=\frac{\lambda}{4\pi \varepsilon_0}\ln(\frac{1-2e^{-\pi |x|/a}\cos(\frac{\pi}{a}(d+y))+e^{-2\pi |x|/a}}{1-2e^{-\pi |x|/a}\cos(\frac{\pi}{a}(d-y))+e^{-2\pi |x|/a}})$$
In Desmos, plotting these three expressions show that they are identical: link to plot The prefactor $\frac{\lambda}{\pi \varepsilon_0}$ has been removed and the variables y and z swapped. However, I have no idea how to show that the potential from the method of images is equivalent to the potential given by solving Laplace's equation (either formula).
My best guess is to divide the polynomial to get an expression of the form $\ln(1+u)$, then Taylor expand the natural log and swap the order of the summations:
$$\Phi(x,y)=\frac{\lambda}{4\pi \varepsilon_0}\sum_{m=1}^{\infty}\frac{(-1)^m}{m}\sum_{n=-\infty}^{\infty}(\frac{2dy-4nad}{x^2+(y-2na-d)^2})^m$$
I have no idea how to simplify the second summation, but I figured that if it could be written as $u^m$ then the problem would be close to solved.