Could anyone advise on this problem?
Let $g(z$) be an analytic function in punctured ball $B(z_1, R) - \{z_1\}$ and let $N$ be a fixed non-negative integer such that $\lim_{z\rightarrow\ z_1}(z- z_1) ^{m}g(z)=0$ $\forall m > N$, and $\lim_{z\rightarrow\ z_1}(z- z_1)^{n}g(z)= \infty$ $\forall $n < N. Determine the type of singularity of $g(z)$ at $z=z_1$.
Thank you.
Here's another version. Define $G(z) = (z - z_1)^ {N+1} g(z)$ when $z \ne z_1$ and $G(z) = 0$ when $z = z_1$. Then $G(z)$ is analytic on the punctured disk and continuous on the disk, so is analytic on the whole disk using Morera's Theorem. Since $G(z_1) = 0$, you see that $H(z) = \displaystyle \frac{G(z)}{z-z_1} = (z - z_1)^ {N} g(z)$$ is also analytic on the whole disk.
In addition, $H(z_1) \ne 0$, because otherwise $\lim \limits _{z \to z_1} (z-z_1)^{N-1}g(z) = \lim \limits _{z \to z_1} \displaystyle \frac{H(z) - 0}{z - z_1}$ would be finite.
Then you can write $g(z) = \displaystyle \frac{H(z)}{(z-z_1)^{N}}$, where $H(z)$ is analytic, $H(z_1) \ne 0$, which shows $z = z_1$ is a pole of order $N$.
So, the answer to your question is that $g$ has a pole of order $N$ at $z_1$. Your hypotheses tell us precisely that $$g(z)=\sum_{k=-N}^\infty a_k(z-z_1)^k$$ with $a_{-N}\ne 0$.
In order for $g$ to have a removable discontinuity at $z=z_1$, $\lim_{z\to z_1}g(z)$ must exist (be a complex number). Equivalently, there must be a function $f$ analytic on $B(z_1,R)$ such that $g(z)=f(z)$ for all $z\in B(z_1,R)\setminus\{z_1\}.$ Equivalently, the Laurent series of $g$ in $B(z_1,R)\setminus\{z_1\}$ must have no negative powers of $(z-z_1).$
In order for $g$ to have a pole at $z=z_1,$ $\lim_{z\to z_1}|f(z)|$ must be infinite. Equivalently, there must be function $f$ analytic in $B(z_1,R)$ and a positive integer $n$ such that $g(z)=\frac{f(z)}{(z-z_1)^n}$ for all $z\in B(z_1,R)\setminus\{z_1\}.$ Equivalently, the Laurent series of $g$ in $B(z_1,R)\setminus\{z_1\}$ must have only finitely many negative powers of $(z-z_1).$
In order for $g$ to have an essential singularity at $z=z_1,$ $\lim_{z\to z_1}|f(z)|$ must fail to exist in any sense. Equivalently, the "equivalently" conditions above fail.
