# Finding Laurent series representation of rational function

I have a question regarding an exercise that is concerned with a Laurent series expansion. This is my solution for a):

Consider the annuli
$$A_1:=\{z \in \mathbb{C} \mid 0<|z-1|<2 \} \\ A_2:=\{z \in \mathbb{C} \mid 0<|z+2i|<2 \}.$$

We have, $$|-2i-1| =|2i+1| =\sqrt{5} >2.$$ So $$1 \notin A_2, \ -2i \notin A_1$$.

Consider $$f(z) =\frac{5}{z^2+(2i-1)z-2i} =\frac{5}{(z+2i)(z-1)}$$ and $$f_1: A_1 \to \mathbb{C}, \ g(z)=\frac{5}{z+2i}=(z-1) f(z) \\ f_2: A_2 \to \mathbb{C}, \ h(z)=\frac{5}{z-1}=(z+2i) f(z).$$

By the above discussion, $$g$$ and $$h$$ are holomorphic on their respective domains as they have no singularities. By the Laurent series theorem we obtain the representations $$f_1(z)=\sum_{n=-\infty}^{\infty} a^{(1)}_n (z-1)^n$$ $$f_2(z)=\sum_{n=-\infty}^{\infty} a^{(2)}_n (z+2i)^n,$$ where $$a^{(1)}_n =\frac{1}{2 \pi i} \int_{|z-1|=\rho_1} \frac{f_1(z)}{(z-1)^{n+1}}$$ $$a^{(2)}_n =\frac{1}{2 \pi i} \int_{|z+2i|=\rho_2} \frac{f_2(z)}{(z+2i)^{n+1}},$$ with $$0<\rho_i<2, \ i=1,2$$. From this we obtain the representations $$f(z) =\frac{f_1(z)}{z-1} =\sum_{n=-\infty}^{\infty} a^{(1)}_n (z-1)^{n-1}$$

$$f(z) =\frac{f_2(z)}{z+2i} =\sum_{n=-\infty}^{\infty} a^{(2)}_n (z+2i)^{n-1}.$$

on $$A_1,A_2$$ respectively. Using the substitution $$k=n-1$$ we obtain

$$f(z)=\sum_{k=-\infty}^{\infty} a^{(1)}_{k+1} (z-1)^k$$

$$f(z) =\sum_{k=-\infty}^{\infty} a^{(2)}_{k+1} (z+2i)^k.$$

By defining $$a_k:=a^{(1)}_{k+1}, b_k:=a^{(2)}_{k+1}$$, we obtain the Laurent Series for $$f$$ around $$1$$ and $$-2i$$. q.e.d.

However, I am not entirely satisfied with this. Considering the wording of b) I think I am supposed to find a more explicit formula for the Laurent Series. But how does one derive a more explicit formula in this case? The only thing I was able to observe is that $$f$$ has poles at the singularities, therefore the Laurent Series expansions should have finite principal part. But I do not see how this helps.

• Do a partial fraction decomposition. One of the summands can be expanded as a Taylor series, the other will be in Laurent series form already. Jun 10, 2021 at 16:29

You have$$f(z)=\frac{1-2 i}{z-1}-\frac{1-2 i}{z+2 i}.$$Note that, if $$|z-1|<\sqrt5$$, then\begin{align}\frac{1-2i}{z+2i}&=\frac{1-2i}{2i+1+(z-1)}\\&=\frac{1-2i}{1+2i}\cdot\frac1{1+\frac{z-1}{1+2i}}\\&=\frac{1-2i}{1+2i}\sum_{n=0}^\infty\frac{(-1)^n}{(1+2i)^n}(z-1)^n,\end{align}since $$\left|\frac{z-1}{1+2i}\right|<1$$. So, near $$1$$, the Laurent series of $$f(z)$$ is$$\frac{1-2i}{z-1}-\sum_{n=0}^\infty\frac{(1-2i)(-1)^n}{(1+2i)^{n+1}}(z-1)^n.$$In particular, $$\operatorname{res}_{z=1}\bigl(f(z)\bigr)=1-2i$$.
Also, you have, near $$-2i$$,\begin{align}\frac{1-2i}{z-1}&=\frac{1-2i}{-1-2i+z+2i}\\&=\frac{-1+2i}{1+2i}\cdot\frac1{1-\frac{z+2i}{1+2i}}\\&=(-1+2i)\sum_{n=0}^\infty\frac{(z+2i)^n}{(1+2i)^{n+1}},\end{align}and so the Laurent series of $$f(z)$$ near $$-2i$$ is$$-\frac{1-2i}{z+2i}+(-1+2i)\sum_{n=0}^\infty\frac{(z+2i)^n}{(1+2i)^{n+1}}.$$
• Is the computation of the Laurent series near $-2i$ actually correct? You seem to have a series of the form $\sum_{n=-\infty}^{\infty} a_n (z \textbf{-} 2i)^n$ which seems to be a series around $2i$. Shouldn't it be a series of the form $\sum_{n=-\infty}^{\infty} a_n (z+2i)^n$? In fact, I got $(1-2i) \sum_{n=0}^{\infty} \frac{-1}{(1+2i)^{n+1}} (z+2i)^n -\frac{1-2i}{z+2i}$ as a result. Jun 10, 2021 at 21:09