I have a question regarding an exercise that is concerned with a Laurent series expansion.

enter image description here

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.

  • 1
    $\begingroup$ 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. $\endgroup$ Jun 10, 2021 at 16:29

1 Answer 1


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$$-\frac{1-2i}{z+2i}+(-1+2i)\sum_{n=0}^\infty\frac{(z+2i)^n}{(1+2i)^{n+1}}.$$

  • $\begingroup$ 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. $\endgroup$
    – Polymorph
    Jun 10, 2021 at 21:09
  • 1
    $\begingroup$ Right you are! Sorry about that. I've edited my answer. $\endgroup$ Jun 10, 2021 at 21:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .