# Principal part of Laurent series when there's multiple singularities

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|.

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.

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.

• Thank you so much, it's way more clear this way :) May 22 at 20:07