Let $a\in \mathbb C$ and $f$ be a complex function defined on $U,$ where $U$ is an open set s.t. $a\in U.$
I hear that the statement below holds, but I'm not sure whether this is correct.
Suppose $a$ is a singularity of $f$.
If there exist $n\in \mathbb N$ and $A\in \mathbb C\setminus \{ 0\}$ s.t. $\displaystyle\lim_{z\to a}\ (z-a)^n f(z)=A,$ then $a$ is a pole of order $n$.
I know that if $f$ has the form $f(z)=\displaystyle\sum_{k=1}^n \dfrac{c_{-k}}{(z-a)^k}+\sum_{k=0}^\infty c_k(z-a)^k$, then $a$ is a pole of order $n$.
So, it's sufficient to investigate whether $f$ has this form when $\displaystyle\lim_{z\to a}\ (z-a)^n f(z)=A\neq 0$.
My try is here.
If $(z-a)^n f(z)$ is regular analytic, then $\displaystyle(z-a)^n f(z)=\sum_{k=0}^\infty c_k (z-a)^k$ so $f(z)=\dfrac{c_0}{(z-a)^n}+\cdots$ and I can see that $a$ is the pole of order $n.$
So I tried to check $(z-a)^n f(z)$ is regular analytic, i.e., $\displaystyle\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$ exists.
I have $\displaystyle\lim_{w\to 0} w^n f(a+w)=A$.
$\displaystyle\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}=\lim_{h\to 0} h^n f(a+h)\dfrac{1}{h^n f(a+h)}\dfrac{f(a+h)-f(a)}{h}$.
This doesn't seem to work. (and the condition $f(a+h)\neq 0$ is needed.)
Thanks for any idea for this statement.