Poles and several singularities

[Ende Jin]

1. In the book, when talking about poles: (see attachment 1)
It declares "there is no harm to assume $|f(z)| > 1 $ in $0 < |z - z_0| < r_0$". But why there is no harm? I mean I understand there exists a small ball around $z_0$ such that $f(z)$ can be very big, however, you see that $g(z) = \frac{1}{f(z)}$, if there is a point in $0 < |z_1 - z_0| < r_0$ s.t. $f(z_1) = 0$, that means I must find a smaller $r_0' < |z_1 - z_0| \le r_0$ ,s.t., $|f(z)| > 1 $ in $0 < |z - z_0| < r_0'$ and go on with this proof. However, that means the decomposition $\frac{H(z)}{(z-z_0)^m} = f(z)$ is only valid in $0 < |z - z_0| < r_0'$, then what about the part $r_0' \le |z - z_0| < r_0$?

2. (Attachment 2) I cannot understand this part: what does repeat mean? I have no idea how to extend the above argument into the situation where there are several poles in the domain (). I can understand how to do it when there are only several removable singularities though.

[Ende Jin]

I suddenly realized that I can plus a polynomial (big enough in the domain) in my question 1.

Although that doesn't help with the question 2.

[Victor Ivrii]

1. If $z_0$ is a pole, then $|f(z)|\to \infty$ as $z\to z_0$ and therefore $\forall M \exists r>0: |z-z_0|<r\implies |f(z)|\ge M$.

2. If $z_0$ is a pole of order $m$ them $f(z)=\sum_{n=-m}^\infty a_n(z-z_0)^n$. Then $f(z)=P((z-z0)^{-1}+g(z)$ with $P((z-z_0)^{-1})=\sum_{n=-m}^{-1}a_n(z-z_0)^n$.

If $f(z)$ is analytic in $D$ except poles $z_1,\ldots, z_N$, then $f(z)=P_1((z-z_1)^{-1} +g(z)$ where $g(z)$ is analytic in $D$ except poles $z_2,\ldots, z_N$. "Repeat" then $g(z)=P_1((z-z_2)^{-1} +h(z)$ where $g(z)$ is analytic in $D$ except poles $z_3,\ldots, z_N$ and so on...

[Ende Jin]

1. I understand that there can be a small ball around $z_0$ that $f$ are very big, since the limit is infinity. However, for example, if f is analytic on $0 <|z-z_0| < r_0$ and we want to find the decomposition on $0 <|z-z_0| < r_0$ . That means I will have to use the limit first so that for some $r_1$, $|f(z)| > 1 \forall 0 < |z  - z_0| < r_1$. After that, we have a decomposition on $0 < |z  - z_0| < r_1$. But that decomposition is only valid in the $ 0 < |z  - z_0| < r_1$ while we need to find a decomposition on $ 0 < |z  - z_0| < r_0$.

I was thinking about adding a polynomial but I am not so sure because what if after adding polynomial, there is still some zero inside the range?

[Victor Ivrii]

Indeed, decomposition is valid only for $0<|z-z_1|<r_1$ but polynomial $P_1((z-z_1)^{-1})$ is defined for all $z\in \mathbb{C}$ and therefore $g(z)=f(z)-P_1(z)$ is defined for all $z\in D$, and analytic there except $z_2,\ldots z_N$.