I have a question about the following passage in Ahlfor's Complex Analysis, page 128. (I conveniently copied it from this question)
For a more detailed discussion of isolated singularities, we consider the conditions (1) $\lim_{z\to a}|z-a|^\alpha|f(z)|=0$, (2) $\lim_{z\to a}|z-a|^\alpha|f(z)|=\infty$, for real values of $\alpha$. If (1) hold for a certain $\alpha$, then it holds for all larger $\alpha$, and hence for some integer $m$. Then $(z-a)^mf(z)$ has a removable singularity and vanishes for $z=a$. Either $f(z)$ is identically zero, in which case (1) holds for all $\alpha$, or $(z-a)^mf(z)$ has a zero of finite order $k$. In the latter case it follows at once that (1) holds for all $\alpha>h=m-k$, while (2) holds for all $\alpha<h$. Assume now that (2) holds for some $\alpha$; then it holds for all smaller $\alpha$, and hence for some integer $n$. The function $(z-a)^nf(z)$ has a pole of finite order $l$, and setting $h=n+l$ we find again that (1) holds for $\alpha>h$ and (2) for $\alpha<h$. The discussion shows that there are three possibilities: (i) condition (1) holds for all $\alpha$, and $f(z)$ vanishes identically; (ii) there exists an integer $h$ such that (1) holds for $\alpha>h$ and (2) for $\alpha<h$; (iii) neither (1) nor (2) holds for any $\alpha$.
I couldn't verify the sentences in bold; where do these statements follow from? The finite Taylor expansion given on page 125?
If $(z-a)^mf(z)$ has a zero of finite order $k$ at $a$, then, near $a$, we have$$(z-a)^mf(z)=a_k(z-a)^k+a_{k+1}(z-a)^{k+1}+\cdots$$and therefore$$f(z)=a_k(z-a)^{-(m-k)}+a_{k+1}(z-a)^{-(m-k)+1}+\cdots$$So, if $\alpha>m-k$, $\lim_{z\to a}\lvert z-\alpha\rvert^\alpha\bigl\lvert f(z)\bigr\rvert=0$, and, if $\alpha<m-k$, $\lim_{z\to a}\lvert z-\alpha\rvert^\alpha\bigl\lvert f(z)\bigr\rvert=\infty$. The other sentence in bold can be proved in a similar way.