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I'm taking a complex variable class and I've recently been introduced to the Laurent series as

$$f(z)=\sum_{-\infty}^{\infty}a_n(z-z_0)^n$$

where $z_0$ is the singularity point and the expression holds for $z \in D^*= \{ z\in \mathbb{C} : |z|<r \} - \{z_0 \} $.

However, further in the subject, I've come across the following theorem:

Given $f(z)=1/Q(z)$ with $Q(z)$ a polinomial of roots $\alpha_1,...,\alpha_r$, then you can rewrite $f$ as the sum of polinomials in $1/(z-\alpha_i)$

Now, the proof starts by considering the principal parts $P_1,...,P_r$ of the function $f$ in each singularity $\alpha_i$.

My problem is that I don't fully understand what it means for a function to have different principal parts for different singularities. As far as I was concerned, the Laurent series was only developed for a single singularity.

I also fail to understand the meaning behind defining a new function as the original one minus all the principal parts, i get that this has to do with "removing singularities" but I can't see why it works (mainly because I don't really understand the mathematical expression for the Laurent series of a function with multiple singularities).

Just to make sure, it's not the theorem itself I'm having trouble with but more with the fact that I don't understand the Laurent series for a function with more than one singularity.

I'd appreciate if someone could help me with this issue.

Thanks in advance and sorry for my poor English.

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1 Answer 1

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Let's recap the proof in bite-sized bits. Feel free to comment on the parts that are unclear.

(1) At each isolated singularity $a_k$, your function $f$ has a Laurent expansion $f(z)=\sum_{j} c_j (z- a_k)^k$ that converges near that point.

(2) Each singularity is a pole (of some finite order $m_k\geq 1$).

(3) We can toss out the terms with positive powers to create a rational function $R_k(z)= \sum_{j=1}^{m_k} c_j (z- a_k)^{-j}$ that decays to zero as $z\to \infty$. This is called the principal part of the singularity at that pole.

(4) Because your function $f(z)$ decays at infinity and has only finitely many poles, $g(z)=f(z)-\sum_k R_k(z)$ also decays at infinity and by construction has only removable singularities.

(5) Thus $g(z)$ is entire and bounded hence constant, and that constant can be evaluated by letting $z$ tend to infinity. Thus $g(z)$ is identically zero by Liouville's Theorem.

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  • $\begingroup$ Thank you so much, it's way more clear this way :) $\endgroup$ Commented May 22, 2023 at 20:07

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