A multiple choice question related to singularities , poles. 1)Let $f(z)=\frac{1}{e^z-1}$ for all z$\in C$ such that $e^z\ne1$ then

a) f is meromorphic
b) the only singularities of f are poles
c) f has infinitely many poles on the imaginary axis
d) each pole of f is simple.

2)For z $\in C$, define $f(z)=\frac{e^z}{e^z-1}$ then

a) f is entire
b) the only singularities of f are poles
c) f has infinitely many poles on the imaginary axis
d) each pole of f is simple.

For 1st example I guess $e^z$  is entire function so option a) is correct. Also option b) is correct.
For 2nd example, also I guess options a) & b) are correct.
Please help me to get correct options.

Hint: Using L'Hospital Calulate lim$_{z\to 2n\pi i}\frac{z-2n\pi i}{e^z-1}$,hence deduce that the given function has simple poles at $z= 2n \pi i$, $\forall n \in \mathbb N$
