Show that $h$ has a pole of order $j$ at $a \in \Omega$

Question

Information: This is a proof verification question, where the OP needs both development how to write formally and also needs verification whether the proof is correct in terms of the logic.

Question: Let $\Omega \subset \mathbb{C}$ is open and connected, $h: \Omega \setminus {a} \rightarrow \mathbb{C}$, where $ a \in \mathbb{C}$, is of the form

$h(z) = \frac{a_j}{(z-a)^j} + ... + \frac{a_1}{z-a} + \phi(z)$, where $\phi : \Omega \rightarrow \mathbb{C}$ is holomorphic, $ j \in \mathbb{N}$, and ${a_1},...,{a_j} \in \mathbb{C}$ with ${a_j} \neq 0$. Show that h has a pole of order $j$ at $a \in \mathbb{C}$.

Proof: We need to show two things for the proof. First, $\lim_{z\to a} \lvert h(z) \rvert = \infty$. And we also show, there exists a unique function $g : \Omega \rightarrow \mathbb{C}$ with $g(a) \neq 0$ s.t. $h(z) = \frac{g(z)}{(z-a)^j}$.

Showing the limit: As $z \rightarrow a$, the principal part goes to infinity. Since every $a_j$ is some coefficient, and denominators force the value to infinity, and also $\phi(z)$ is holomorphic on $\Omega$ which implies that it is bounded near $a$, we know that limit goes to infinity.

Finding g: Consider $g(z) = (z-a)^jh(z)$. Since both $(z-a)^j$ and $h(z)$ are holomorphic, simply $g(z)$ is holomorphic too. Also, $g(a) \neq 0$. Therefore, we can write $h(z) = \frac{g(z)}{(z-a)^j}$. This $j$ is called the order of pole, by definition.

Answer 1

Comments on formal writing and the proof:

1. Question: You should mention that $\Omega \subset \Bbb C$ is open and $a\in \Omega$.
2. Proof: We want $g$ to be holomorphic - just any function will not do.
3. Proof 2: Isn't $\lim_{z\to a} |h(z)| = \infty$ unnecessary? If you find the required holomorphic function $g: \Omega\to \Bbb C$ with $g(a) \ne 0$ such that $h(z) = g(z)/(z-a)^j$, doesn't $\lim_{z\to a} |h(z)| = \infty$ follow from the boundedness of $g$ on some compact set containing $a$?
4. Showing the limit: "since every $a_j$ is some coefficient" can be replaced by something along the lines of "since every $a_j \in \Bbb C$ is fixed, the $a_j$'s are bounded" - since that's what you really use for the limit.