There is a well-known identity, which holds in a distribution sense: $$\frac{1}{x+ia} = P\frac 1x -\pi\delta(x).$$ Here, $\lim_{a\to 0^+}$ is understood, $P\frac 1x$ denotes the Cauchy principal value distribution.
Since $\tanh x \approx x$ for small $x$, one may suspect that $$\frac{1}{\tanh(x+ia)} = P\frac{1}{\tanh(x)} -\pi\delta(x).$$ Is this identity true?
I think this is true. Considering the difference between the second and the first identity, enough to show that: $$\frac{1}{\tanh(x+ia)} - \frac{1}{x+ia} = P\frac{1}{\tanh x}-P\frac 1x.$$ However, the function $1/\tanh z - 1/z$ is regular near $z=0$. Hence, we can safely take $a\to 0$ and remove the $P$ symbol. Is my reasoning correct?