Solution
= Solution
{c}
The denominator vanishes exactly when $e^z=-1$, namely at $z=(2k+1)\pi i$, $k\in\mathbb Z$. Its <derivative> there is $e^z=-1\ne0$, so each such zero is simple and the numerator does not cancel it. Hence
$$
\boxed{\frac1{1+e^z}\text{ is holomorphic on }\mathbb C\setminus\{(2k+1)\pi i:k\in\mathbb Z\},}
$$
and every omitted point is a <simple pole>.