In the book that I'm reading it was done as follows:
Notice that the denominator is zero when $z=\pi i$ (That is a singularity).
Now observe:
$(z-\pi i)f(z)=\frac{(z-\pi i)e^{az}}{e^z-e^{\pi i}}=\frac{e^{az}}{\frac{e^z-e^{\pi i}}{z-\pi i}}$.
By considering the limit $z\to\pi i$ we have
$\lim\limits_{z\to\pi i}(z-\pi i)f(z)=e^{a\pi i}\frac{1}{e^{\pi i}}$. Derivative of $e^z=e^z$.
Thus it is a simple pole.
And my question is just because we have a value for the $\lim\limits_{z\to\pi i}(z-\pi i)f(z)$ can we say that is is a simple pole.
Because as far as I have understood, if it is a simple pole then we can use $\lim\limits_{z\to\pi i}(z-\pi i)f(z)$ to find the residue.
But does the other way holds true?
Please correct me If I have understood the theory incorrectly
If $c\neq 0$, $e^z+c$ only has simple zeroes. A multiple zero would be a root of $\frac{d}{dz}(e^z+c)=e^z$ too, and that clearly leads to a contradiction. If follows that $\pi i$ is a simple pole of $\frac{e^{az}}{e^z+1}$ and
$$\operatorname*{Res}_{z=\pi i}\frac{e^{az}}{e^z+1} = \lim_{z\to \pi i}\frac{(z-\pi i)e^{az}}{e^z+1}\stackrel{\text{De l'Hopital}}{=} \lim_{z\to \pi i}\frac{e^{az}}{e^z}=-e^{\pi i a}.$$