This is a problem from 'Mathematical Methods in Physical Sciences' by Mary L. Boas. The question is from Chapter 13, Page 627, Question 6. The problem is as follows:
Show that the series $\begin{align*} T(x,y)=\sum_{n=1,3,5,\cdots}\frac{400}{\pi}\frac{1}{n}e^{\frac{-n\pi y}{10}}\sin\left(\frac{n \pi x}{10}\right) \end{align*}$ can be summed to get $\begin{align*} T=\frac{200}{\pi}\tan^{-1}\left(\frac{\sin(\pi x/10)}{\sinh(\pi y/10)}\right) \end{align*}$
Using the hints, I will show what I have done till now:
My attempt:
$\frac{1}{n}e^{\frac{-n\pi y}{10}}\sin(\frac{n \pi x}{10})=\Im\left[ \frac{\left(e^{\frac{- \pi y}{10}+\frac{i\pi x}{10}}\right)^n}{n}\right]$
Now let $z=e^{\frac{- \pi y}{10}+\frac{i\pi x}{10}}$. Now using $\ln(1+z)$ and $\ln(1-z)$ we have $\ln\left(\frac{1+z}{1-z}\right)=2\left[z+\frac{z^3}{3}+\frac{z^5}{5}+\cdots\right]$
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Then the sum $\begin{align*} T(x,y)=\sum_{n=1,3,5,\cdots}\frac{400}{\pi}\frac{1}{n}e^{\frac{-n\pi y}{10}}\sin\left(\frac{n \pi x}{10}\right) =\Im\left[\frac{400}{\pi}\sum_{n=1,3,5,\cdots}\frac{z^n}{n}\right]=\Im \left[\frac{200}{\pi}\ln\left(\frac{1+z}{1-z}\right)\right] \end{align*}$
How do I proceed from here to get the required result. Thank you.
Best Answer
Going by comments by Gary, I have found the solution, and I will write it here.
$\ln(z)=\ln(re^{i\theta})=\text{Ln}\;r+\ln\;e^{i\theta}=\text{Ln}\;r+i\theta$.
Thus, we have: $\begin{align*}\Im \left[\frac{200}{\pi}\ln\left(\frac{1+z}{1-z}\right)\right]=\frac{200}{\pi}\arg\left(\frac{1+z}{1-z}\right) \end{align*}$
Now, $\arg\left(\frac{1+z}{1-z}\right)=\arg\left(\frac{(1+x)+iy}{(1-x)-iy}\right)$ (multiplying and dividing by conjugate $(1-x)+iy$, to get:$$\arg\left(\frac{1+z}{1-z}\right)=\tan^{-1}\left(\frac{2y}{1-(x^2+y^2)}\right)$$ In our case, we have $x=e^{\frac{-\pi y}{10}}\cos(\frac{\pi x}{10})$ and $y=e^{\frac{-\pi y}{10}}\sin(\frac{\pi x}{10})$
Hence, we get :$$\arg\left(\frac{1+z}{1-z}\right)=\tan^{-1}\left(\frac{2e^{\frac{-\pi y}{10}}\sin(\frac{\pi x}{10})}{1-e^{\frac{-2\pi y}{10}}}\right)=\tan^{-1}\left(\frac{\sin(\frac{\pi x}{10})}{\frac{e^{\frac{\pi y}{10}}-\;e^{\frac{-\pi y}{10}}}{2}}\right)=\tan^{-1}\left(\frac{\sin(\pi x/10)}{\sinh(\pi y/10)}\right)$$
Hence, we have $\begin{align*}T(x,y)=\frac{200}{\pi}\tan^{-1}\left(\frac{\sin(\pi x/10)}{\sinh(\pi y/10)}\right) \end{align*}$