In a (Partial Differential Equations / Laplace Equation) , I try to solving a problem of Laplace eq. by using separation of variables method.
I usually using the rule : if $e^{2 \sqrt{k} b} = 1$, then I have: $2\sqrt{k} b = 2ni\pi$.
Now in my problem I have : $e^{2 \sqrt{k}\pi} = 1$ Can I use the same rule which lead to cancel the $\pi$ ?
On
You're using Euler's formula for complex numbers
$$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$
We know that $\cos(2n\pi)=1, \sin(2n\pi)=0$, for $n\in \Bbb Z$ so:
$$n\in \Bbb Z \implies e^{2ni\pi}=1$$
You have, therefore, in the first part of your question:
$$2b\sqrt k =2ni\pi\to b\sqrt k=ni\pi\to k=-(\frac{n\pi}{b})^2$$
This CAN be applied to part b as well:
$$2\pi\sqrt k=2ni\pi\to\sqrt k=ni\to k=-n^2$$
Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2\sqrt{k}\pi} = 1$, then you have $2\sqrt{k}\pi = 2ni\pi$, and hence $\sqrt{k} = ni$, for some integer $n$.