Classification of infinity as a singular point.

Question

i have the followings problems, i must find and classify the singular points of the followings functions:

$a) \frac{e^{z}}{1+z^{2}}\\$.

$b) \frac{e^{z}}{z(1-e^{-z})}$.

$c) \frac{1}{z^{3}(2-\cos{z})}$.

$d) e^{\frac{z}{1-z}}$.

$e) \cot{z}- \frac{1}{z}$.

$f) \cot{z}- \frac{2}{z}$.

$g) \sin{\frac{1}{1-z}}$.

I have classified the obvious points and they have reviewed it, but they tell me that i must classify infinity as well, but i do not know how to do it, i would greatly appreciate your help please.

Answer 1

Not completely sure if this is rigorous, but my approach would be to make the substitution $ w = 1/z $, and rearrange the functions to be in terms of $w$. Classifying $z = ∞$ will then be equivalent to classifying $w = 0$.

Answer 2

The approach suggested by @user872642 does work, and indeed works well for many functions. However, for some function other methods may lead to simpler calculations. For example, you can look at the behaviour of the function near infinity, i.e. for large $|z|$. The following rules apply:

1. If the function is bounded near $\infty$, it is said to have a removable singularity in $\infty$.
2. If $|f(z)| \rightarrow \infty$ as $|z| \rightarrow 0$, the function is said to have a pole. If you want to find the order of that pole, you might have to do the substitution $w=\frac{1}{z}$.

3. If the function neither vanishes nor blows up as $|z| \rightarrow \infty$, it has an essential singularity in $\infty$.

You can also look at the Laurent series of $f(\frac{1}{z})$, which is what is sometimes (for example by WolframAlpha) referred to as the Laurent expansion at $\infty$.