Prove that If singular part of Laurent series has infinite many terms, then $\lim_{z\to z_0}(z-z_0)^mf(z)$ doesn't exist for all nautural number $m$.

Question

Given $f$ an analytic function in open $D \subset \mathbb C$, $z_0$ is an isolated singularity defined as $B(z_0;r)\backslash\{z_0\} \subset D$, then know that $f$ can be written as an expansion of laurent series $$f(z)=\sum_{k=1}^{\infty}b_{-k}(z-z_0)^{-k}+\sum_{k=0}^{\infty}a_k(z-z_0)^k$$ in $B(z_0;r)\backslash\{z_0\}$, If there are infinite $b_{-k}\neq 0$, prove that $$\nexists\lim_{z\to z_0}(z-z_0)^mf(z)$$ for all nautural number $m$.

My attempt: Assume $\exists m: \lim (z-z_0)^m f(z)=\lim (z-z_0)^m (\sum_{k=1}^{\infty}b_{-k}(z-z_0)^{-k}+\sum_{k=0}^{\infty}a_k(z-z_0)^k)=c$, then $\lim(z-z_0)^m \sum_{k=0}^{\infty}a_k(z-z_0)^k=\lim(z-z_0)^m\lim \sum_{k=0}^{\infty}a_k(z-z_0)^k=0\cdot0=0$ as $\sum_{k=0}^{\infty}a_k(z-z_0)^k$ continuous in $B(z_0;r)$, then $\lim (z-z_0)^m \sum_{k=1}^{\infty}b_{-k}(z-z_0)^{-k}=$ $$\lim_{z\to z_0} \sum_{k=m}^{\infty}b_{-k}(z-z_0)^{-k+m}=c$$ What can I do next?

Edit: In my comment it’s $a_{-k}=0$

Edit2: I follow these definitions for classifications of isolated singularities:

Removable singularity: No singular part of laurant series. Pole of order m: singular part with finite $a_{-k}\neq 0$, $a_{-m}\neq 0$ is the last term $\neq 0$ and $a_{-k}=0$ $\forall k>m$. Essential singularity: singular part with infinite many $a_{-k}\neq 0$.

and I try to use these definitions to imply other than using other definitions.

Answer 1

In Fundamentals of Complex Analysis Engineering, Science, and Mathematics $3/E$ pg $284$

Theorem 18

$f$ has an isolated singularity at $z_0$

$(1)$ $z_0$ removable singularity iff $|f|$ is bounded near $z_0$

$(2)$ $z_0$ pole iff $|f| \to \infty$ as $z \to z_0$

$(3)$ $z_0$ essential singularity iff $|f|$ neither above.

Proof:

For $(1)$, $(2)$:

$\implies$: Trivial by using the definitions mentioned in Edit2.

$\impliedby$:

$(1)$

In Fundamentals of Complex Analysis Engineering, Science, and Mathematics $3/E$ pg $286$

13. Let $f (z)$ have an isolated singularity at $z_0$ and suppose that $f (z)$ is bounded in some punctured neighborhood of $z_0$. Prove directly from the integral formula forthe Laurent coefficients that $a_{− j} = 0$ for all $j = 1, 2,...$; that is, $f (z)$ must have a removable singularity at $z_0$

In Saff, Snider - Complex Analysis Solutions Manual (3rd Ed.) pg $5-33$

Suppose $|f(z)|<M$ for $|z-z_0|<\rho<R$ and define $C$ to be the positively oriented circle $|z-z_0|=\rho$.

Then $$|a_{-j}|=|\frac{1}{2\pi i}\int_{C}\frac{f(w)}{(w-z_0)^{-j+1}}\quad dw|\le\frac{1}{2\pi}\frac{M}{\rho^{-j+1}}2\pi\rho=\rho^j M$$

Letting $\rho\to 0$ shows that $a_{-j}=0$.

$(2)$

In https://www.math.uci.edu/~ndonalds/math147/6residues.pdf pg$7$

Definition 6.10. Suppose $f(z)$ is analytic at $z_0$ and $f(z_0) = 0$. We say that $z_0$ is a zero of order $m$ if $f^{(m)}(z0)$ is the first non-zero derivative.

In pg$9$

Theorem 6.15. Let $f(z)$ be analytic at $z_0$ and $g(z) = \frac{1}{f(z)}$.

Then, at $z_0$, $f(z)$ has a zero of order $m$ iff $g(z)$ has a pole of order $m$.

Hint(pg $13$): use this theorem to solve $\impliedby$.

Proof:

$g\to \infty\implies f=1/g\to 0$ as $z\to z_0$.

So $f$ is bounded in deleted neighborhood $\implies$ $f$ has a removable singularity at $z_0$ $\implies$ $f$ has a Laurant series $\Sigma_{j=0}^{\infty}a_j(z-z_0)^j$ (Power series).

$g$ is analytic in deleted neighborhood with radius $R$ $\implies$ $f$ is also analytic $\implies$ radius of convergence of power series $\ge R$.

As power series is continuous at $z_0$, suppose $f(z_0)=0$, so $a_0=0$.

Assume that all $a_k=0$, then $f=0$ in the neighborhood, then $g$ doesn't exist in deleted neighborhood, contradiction.

So $\exists a_k\neq 0$, assume there is the $a_l\neq 0$ satisfies $a_k=0 \forall k < l$, so $f^{(l)}(z_0)=l!a_l\neq 0$, f has a zero of order $l$, so $g$ has a pole, the $\impliedby$ in $(2)$ has been proved.

As $(1)$, $(2)$ have been proved, so $(3)$ is trivial.

$(3)$ means that an essential singularity,$\lim_{z\to z_0}$ doesn't exist in $C \cup \{\infty\}$, also, if $f$ has an essential singularity, after calculation we know $(z-z_0)^mf(z)$ also has an essential singularity, so the question has been proved.

Answer 2

If $\lim (z-z_0)^m f(z)$ exists then $(z-z_0)^m f(z)$ has a removable singularity at $z_0$, so there is a power series expansion $(z-z_0)^m f(z)=\sum\limits_{n=0}^{\infty} a_n(z-z_0)^{n}$. This gives a Laurent series expansion $f(z)=\sum\limits_{n=0}^{\infty} a_n(z-z_0)^{n-m}=\sum\limits_{n=-m}^{\infty} a_{n+m}(z-z_0)^{n}$. (Recall that Laurent series expansion is unique). Hence, there are only finitely many terms in negative powers of $z-z_0$.