Learn Analyzing Functions with These Examples

Types of Singularities for singularities of the complex function $\frac{1-<!--\; -\; -->e^z}{1+e^z}$
I have two questions:
Question 1
For Functions with a holomorphic numerator and polynomial denominator, can I simply find the Taylor Series of the numerator in the neighbourhood of the first Singularity, and determine the type in the neighbourhood? And do so for each singularity?
Example
Given $f(z)=\frac{e^z}{z^2+1}$ with Singularities at $z=\pm <!--\; \pm \; -->i$. Let us check the type of z=i first. Since $e^z\in <!--\; \in \; -->H(\mathbb{C})$, we can use the taylor expansion in the neighbourhood of i, ie. $B_{\mathrm{ϵ<!--\; ϵ\; -->}}(i)$ where $0<\mathrm{ϵ<!--\; ϵ\; -->}<2$ so that $-<!--\; -\; -->i\notin <!--\; \notin \; -->B_{\mathrm{ϵ<!--\; ϵ\; -->}}(i)$. Then
$f(z)=\frac{1}{(z-<!--\; -\; -->i)(z+i)}\underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{e^i}{k!}(z-<!--\; -\; -->i{)}^k$
$=\frac{1}{z+i}\underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{e^i}{k!}(z-<!--\; -\; -->i{)}^{k-<!--\; -\; -->1}$
$=\frac{1}{z+i}(\frac{e^i}{z-<!--\; -\; -->i}+\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{e^i}{k!}(z-<!--\; -\; -->i{)}^{k-<!--\; -\; -->1})$
So i has a Pole of first order. Similarily we can show that −i also is a Pole of first order.
Is that right?
Question 2 How do you go about functions like $f(z)=\frac{1-<!--\; -\; -->e^z}{1+e^z},|<!--\; |\; -->z|<!--\; |\; --><4$?
I've found the Singularities, namely $z=\pm <!--\; \pm \; -->i\mathrm{\pi <!--\; \pi \; -->}$, but have no idea how to proceed, because I can't find a way to use a similar approach to the above example...!

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