If $|f(z)| \leq C|z|^M$ , then $z = 0$ is either removable or a pole. Let $f$ be a holomorphic function on the punctured plane $\mathbb{C}− \{0\}$. Assume that
there exist a positive constant $C$ and a real constant $M$ such that $|f(z)| \leq C|z|^M$ for $0 \lt |z| \lt \frac{1}{2}$.Show that the singularity at $z = 0$ is either removable or a pole.
This question has a answer here.but I couldn't understand the hint. I want a more detailed solution. Any help is appreciated
 A: Let $f$ be a function as in the description i.e. $f$ is holomorphic in $\mathbb C \setminus \{0\}$ and $|f(z)| \leq C |z|^M$ for $0<|z|<\frac 12$. This answer requires no knowledge of Laurent series, and appears to be a more natural fit because of its basic nature.

We will use the following theorem :

Theorem : (10.20, Rudin's Real and Complex Analysis) Suppose that $\Omega$ is a domain and $a \in \Omega$. Let $f$ be holomorphic in $\Omega \setminus \{a\}$ and suppose that there exists $r>0$ such that $f$ is bounded in the set $\{z : |z-a| < r\}$. Then, $f$ has a removable singularity at $a$.

The proof of this is as follows : let $h$ be defined on $\Omega$ by $h(a) = 0$ and $h(z) = (z-a)^2f(z)$ in $\Omega \setminus \{a\}$. Then, note that $h$ is holomorphic in $\Omega \setminus \{a\}$. Furthermore,
$$
\lim_{z' \to 0}\frac{h(a+z') - h(a)}{|z'|} = \lim_{z' \to 0} \frac{(z')^2f(z')}{|z'|}
$$
Let $|f| \leq M$ in $\{z : |z-a|<r\}$, where $M$ is a bounding constant. Then, we have $$0 \leq \left|\frac{(z')^2f(z')}{|z'|}\right| \leq M |z'|$$
for all $|z'|<r$. By the squeeze theorem, it follows that $$h'(a) = \lim_{z' \to 0} \frac{(z')^2f(z')}{|z'|} = 0$$
Therefore, $h$ is holomorphic in $\Omega$. By Cauchy's theorem, $h$ admits a power series expansion in $\Omega$ of the form $$
h(z) = \sum_{n=0}^{\infty} c_n(z-a)^n
$$
However, note that $h(a) = 0$ and $h'(a) = 0$ by definition. Therefore, the power series above starts from $2$ (because $c_0 = h(a)$ and $c_1 = h'(a)$, when one observes the proof of Cauchy's theorem). That is, $$
h(z) = \sum_{n=2}^{\infty} c_n(z-a)^n
$$
Therefore, if $g(z) = \frac{h(z)}{(z-a)^2}$, then $g(z) = \sum_{n=0}^{\infty} c_{n+2}(z-a)^n$ is holomorphic in $\Omega$ because it admits a power series expansion, and also note that $g=f$ on $\Omega \setminus \{a\}$. This shows that $f$ has a removable singularity at $\{a\}$, as desired.

To use this theorem, we take $r = \frac 12$ and consider the function $g(z) = z^N f(z)$ for a positive integer $N$. Clearly, $g$ has an isolated singularity at the point $0$.
Then, note that in the region $|z|<\frac 12$, $$
|g(z)| = |z^Nf(z)| \leq C|z^N||z^M| \leq C|z^{N+M}|
$$
Now, suppose that $N$ is chosen so that $N+M \geq 0$ ($0$ is included because $|z|^0 = 1$ is also a monotone decreasing function in the modulus). Then, for all $|z|<\frac 12$, we have $$
C |z|^{N+M} \leq \frac{C}{2^{N+M}}
$$
i.e. $g(z)$ is a bounded function in $|z|<\frac 12$ with $0$ as an isolated singularity. This proves that $0$ is a removable singularity of $g(z)$.
Therefore, $z^Nf(z)$ has a removable singularity at $0$ for all $N \geq -M$ an integer. However, note that $z^Nf(z)$ has a removable singularity at $a$ if and only if the order of the pole at $a$ is at most $N$.
Consequently, the following results are obtained :

If $M \geq 0$, then $f(z)$ has a removable singularity at $a$.


If $M < 0$, then $f(z)$ has a pole at $a$ of order at most $N = \lceil -M\rceil$, or has a removable singularity at $a$.

To see why it is at most $N = \lceil M\rceil$ and not exact or equal, take $f(z) = z^{-1}$.Then $f$ has a pole of order $1$ at $0$.
However, for any $M<-1$, it is true that $|f(z)| < C|z|^M$ for some constant $C$ in $|z|<\frac 12$. Indeed, $|z^{-1}| \leq C |z|^{M}$ in a neighbourhood of $0$ if and only if $|z|^{-M-1}$ is upper bounded near $0$, which happens precisely when $-M-1>0$ i.e. $M<-1$.
This is also the case if one tries a removable singularity. Hence, the result is obtained without the usage of Laurent series.
