Show that $\frac{1}{f(z)}$ has a removable singularity at $z_0$

Question

Working on some exam practice questions and trying to get a better understanding of the solutions.

Suppose that $f$ is an analytic function on an open set $D$ with isolated singularity at $z_o$ and supposed that we have already proven that $|f(z)| \rightarrow \infty$ as $z \rightarrow z_0$

The 2nd part of the question says:

Show that $\frac{1}{f(z)}$ has a removable singularity at $z_0$, and that there exists $m \ge 1$ and an analytic function $h$ such that $h(z_0) \ne 0$ and $$\frac{1}{f(z)} = (z-z_0)^mh(z)$$

So the solution says:

Since $|f(z)| \rightarrow \infty$ as $z \rightarrow z_0$, $\exists \space \delta > 0$ such that $|f(z)|> 1$ whenever $0 < |z-z_0|< \delta$. So $\frac{1}{f(z)}$ is well defined and analytic on $B(z_0,\delta) \backslash {z_0}$ and $z_0$ is an isolated singularity.

My question at this point is this: Why does that make $1/f$ analytic? I'm failing to see why that necessarily means we can represent $f(z)$ with a power series for all $ z \in B(z_0,\delta) $?

The solution goes on to say:

Since $$\lim_{z \rightarrow z_0} \frac{1}{f(z)} = 0 \space \space (\star) $$ $z_0$ is an isolated singularity. Why is this the case? Is this because we know $z_0$ is an isolated singularity and since the limit is $0$ we know it's not a pole and not an essential singularity (as the limit exists)?

Therefore it has a power series of the form $\sum_{k=0}^{\infty}a_k (z-z_0)^{k} $ Finally, it says (I don't understand this part), due to $(\star)$, there exists $m \ge 1$ such that $a_k = 0$ for all $k = 1,2,...,m-1$, so therefore: $$\frac{1}{f(z)} = \sum_{k=m}^{\infty}a_k (z-z_0)^{k} = (z-z_0)^{m}\sum_{k=m}^{\infty}a_k (z-z_0)^{k-m} = (z-z_0)^mh(z)$$ Then $h$ is analytic on $B(z_0,\delta)$ and $h(z_0)=a_m \ne 0$.

How do we get from $(\star)$ to the next part? I am failing to see the connection there, could someone help me with explanation?

Many thanks, and any help would be greatly appreciated!

Answer 1

First question: A function is (complex) analytic if and only if it is holomorphic. It is an easy exercise to see that if $f$ is complex differentiable in $w$ and $f(w) \neq 0$, then $1/f$ is also complex differentiable in $w$ and $(1/f)'(w) = -f'(w)/f(w)^2$. So we know that $1/f$ is holomorphic in $B(z_0,\delta)\setminus \{z_0\}$, hence analytic.

Second question: It is not correct that $z_0$ is an isolated singularity of $1/f$ since $\lim\limits_{z\to z_0} \frac{1}{f(z)} = 0$. It is an isolated singularity since $1/f$ is holomorphic in $B(z_0,\delta)\setminus\{z_0\}$. The limit says that $z_0$ is a removable singularity, and the value which removes the singularity is $0$, so

$$g(z) = \begin{cases}\frac{1}{f(z)} &, z \neq z_0\\ 0 &, z = z_0 \end{cases}$$

is holomorphic on $B(z_0,\delta)$.

Third question: Since $g(z_0) = 0$, the power series representation of $g$ about $z_0$ has a constant term zero, hence

$$g(z) = \sum_{k=1}^\infty a_k (z-z_0)^k.$$

since $g$ does not vanish identically, not all coefficients are $0$, and if we let $m = \min \{ k \in\mathbb{N} : a_k \neq 0\}$, we have

$$g(z) = \sum_{k=m}^\infty a_k (z-z_0)^k$$

with $a_m \neq 0$. Then

$$h(z) = \sum_{k=m}^\infty a_k (z-z_0)^{k-m} = \sum_{r=0}^\infty a_{r+m}(z-z_0)^r$$

is holomorphic in $B(z_0,\delta)$ with $h(z_0) = a_m \neq 0$, and evidently $g(z) = (z-z_0)^m\cdot h(z)$. Since $g$ has no zero in $B(z_0,\delta)\setminus\{z_0\}$, $h$ does not vanish anywhere in $B(z_0,\delta)$, so $\tilde{h}(z) = \frac{1}{h(z)}$ is holomorphic on $B(z_0,\delta)$, and we have

$$f(z) = (z-z_0)^{-m}\cdot \tilde{h}(z)$$

on $B(z_0,\delta)\setminus\{z_0\}$.

Answer 2

Everything here are answered with standard theorem, so if you find that unsatisfying, you might need to give more information on which theorem have you studied. For example, complex differentiable imply analytic is an obvious way to answer your first question, but it sounds like you were not aware of it at all.

$f(z)$ is complex differentiable, so $\frac{1}{f(z)}$ is complex differentiable, and so is analytic.

To show that $z_{0}$ is an isolated singularity (note that showing that it is an isolated singularity does not involve ruling out essential singularity and pole-you do that when you want to show that it is a removable singularity): $z_{0}$ is an singularity of $f(z)$ and so must be a singularity of $\frac{1}{f(z)}$. To show that it is isolated, simply need to check that $f(z)\not=0$ in a punctured neighbourhood of $z_{0}$, which is indeed the case since $\lim\limits_{z\rightarrow z_{0}}|f(z)|=\infty$. This is because the only way $\frac{1}{f(z)}$ could have singularity are when $f(z)$ have singularity or when $f(z)=0$, and we already know $z_{0}$ is isolated singularity for $f(z)$ so there is a punctured neighbourhood where $f(z)$ have no singularity.

To show that $z_{0}$ is a removable singularity: as said above, limit exist. This is just direct application of Riemann's theorem on removable singularity (limit exist, so $\frac{1}{f(z)}$ is continuously extendable).

Laurent series principle part must be $0$ since this is a removable singularity. So think about the value of $a_{0}$ in the series. $a_{0}=\lim\limits_{z\rightarrow z_{0}}\frac{1}{f(z)}=0$. You know that the series expansion have at least 1 non-zero coefficient (since $\frac{1}{f(z)}$ cannot be constant $0$). Once you know that $a_{0}$ is $0$, you know that there is a minimum $m$ such that $a_{m}\not=0$ and $m\geq 1$. So all term in the series have a factor of $(z-z_{0})^{m}$. Once we factor that one out, we are left with $h(z)$ which have a constant non-zero term in the series expansion (that is $a_{m}$). $h$ is analytic by definition because it is a power series.