My complex analysis textbook stated the following proposition:
Let $a$ be an isolated singularity of $f$
If $\lim_{z\to a}(z-a)f(z)=0$, then $a$ is a removable singularity
If there exists a number $m \in \mathbb{N}$ such that: $\lim_{z\to a}(z-a)^mf(z) \neq 0$, then $a$ is a pole of order $m$
My questions are:
1 - Imagine that we have a function $f$ such that $\lim_{z\to a}(z-a)f(z)=0$. Would that mean that $ \nexists m\in \mathbb{N}: \lim_{z\to a}(z-a)^mf(z) \neq 0$?
2 - Imagine that we have a funcion $f$ such that $\exists m\in \mathbb{N}: \lim_{z\to a}(z-a)^mf(z) \neq 0$, would that mean that: $\forall k \in \mathbb{N} \setminus \{m\},\lim_{z\to a}(z-a)^k f(z) = 0$?
This makes sence because the order of the pole is only one number, so there can only exist one number such that $\lim_{z\to a}(z-a)^mf(z) \neq 0$
3 - Imagine that we have a function $f$ such that $ \exists m\in \mathbb{N}: \lim_{z\to a}(z-a)^mf(z) \neq 0$. This would mean that $a$ is a pole of order $m$.
Now let's consider the limit $\lim_{z\to a}(z-a)^k f(z)$, with $k \in \mathbb{N}$ such that $k \neq m$. We have that either $\lim_{z\to a}(z-a)^k f(z)=0$ (1) or $\lim_{z\to a}(z-a)^k f(z) \neq 0$ (2).
If (2) is true there exists two different number such that $\lim_{z\to a}(z-a)^mf(z) \neq 0$ so with one of them is the order of the pole?
If (2) is false and (1) is true because there exists only one number such that $\lim_{z\to a}(z-a)^mf(z) \neq 0$, if we consider $k = 1$, this would imply that $\lim_{z\to a}(z-a)f(z)=0$ and this would mean that $a$ is a removable singularity because of the initial proposition. But we also have $ \exists m\in \mathbb{N}: \lim_{z\to a}(z-a)^mf(z) \neq 0$ because of the way I defined $f$, so, again because of the proposition, $a$ is a pole of order $m$. This seams like a paradox So witch one is right?
2 in the first paragraph means implicitly that the limit is finite and non-zero; for a pole the limit is infinite for all $k<m$ and zero for all $k>m$, so 2 in the second paragraph is false as noted - in general, there are actually 4 choices - zero limit, finite non-zero limit, infinite limit or no limit (this last being precluded in a pole or removable singularity, but happening in an essential singularity); there is no paradox just a misunderstanding of what the notions involved mean
personally I think that the above definitions are way too complicated and they are properly expressed as theorems or lemmas etc
The standard definitions for an isolated singularity are:
$a$ is a removable singularity if $f$ is bounded in a neighborhood of it (and the result is that there is a finite value $c$ for which $f(z) \to c$ and we call that $f(a)$, and then $f$ is holomorphic near $a$ too, while $(z-a)^mf(z) \to 0, m \ge 1$
$a$ is a pole if $f(z) \to \infty, z \to a$ (and the result is that there is an unique integer $m \ge 1$ for which $(z-a)^mf(z) \to c \ne 0, c \ne \infty$ etc)
$a$ is an essential singularity if the limit of $f(z)$ doesn't exist when $z \to a$ and then there are sequences $z_{mk} \to a, (z_{mk}-a)^mf(z_{mk}) \to \infty, k \to \infty$ for all $m \ge 0$, as well as sequences for which $f$ is bounded - actually by a more general theorem, all but at most one complex value, are taken infinitely many times in a neighborhood of $a$, though the first result about $f$ having sequences on which it grows faster than any rational function near $a$ is also important