Types of Singularities for singularities of the complex function $\frac{1 - e^z}{1 + e^z}$

Question

I have a few questions on how to determine the types of Singularities, given there are more than 1 Singularity.

Unfortunately, a lot of other posts similar to this one has answers using the Residuum of Complex functions (as well as using limits to determine the type), but we haven't covered them in my Complex Analysis course, so I am a bit reluctant to use something we didn't prove.

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 $z=\pm i$. Let us check the type of $z=i$ first. Since $e^z \in H(\mathbb C)$, we can use the taylor expansion in the neighbourhood of $i$, ie. $B_{\epsilon}(i)$ where $0<\epsilon<2$ so that $-i \notin B_{\epsilon}(i)$. Then $$f(z)=\frac{1}{(z-i)(z+i)}\sum_{k=0}^{\infty}\frac{e^i }{k!}(z-i)^k$$ $$=\frac{1}{z+i}\sum_{k=0}^{\infty}\frac{e^i }{k!}(z-i)^{k-1}$$ $$=\frac{1}{z+i}(\frac{e^i}{z-i} + \sum_{k=1}^{\infty}\frac{e^i}{k!}(z-i)^{k-1})$$ 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 $f(z)= \frac{1-e^z}{1+e^z}, \lvert{z}\rvert< 4$?

I've found the Singularities, namely $z=\pm i \pi$, but have no idea how to proceed, because I can't find a way to use a similar approach to the above example...

Thank you folks for your help in advance!

Answer 1

Your approach to the first question is a good one. Regarding the second question, here is an alternative approach that can also be applied to situations such as the first question.

Let's say the function $f(z) = \frac{g(z)}{h(z)}$ has an isolated singularity at $z = z_0$ because $h(z_0) = 0$. Assume that $g$ and $h$ are analytic in a neighborhood of $z_0$. First expand both $g$ and $h$ into power series about $z_0$, say $g(z) = \alpha_k(z-z_0)^k + \alpha_{k+1}(z-z_0)^{k+1} + \dots$ and $h(z) = \beta_m(z-z_0)^m + \beta{m+1}(z-z_0)^{m+1} + \dots$.

If $k \ge m$, this is a removable singularity which is removed by setting $f(z_0) = 0$ in case $k > m$ and $f(z_0) = \frac{\alpha_k}{\beta_k}$ in case $k = m$ (just like in L'Hopital's rule).

If $k < m$, this is a pole of order $m - k$.

Check that this approach gives you the correct answer for question 1, make up another question like this and answer it for yourself, and then apply it to the second question.