# How to determine the order of the poles of a complex function using its Laurent series?

I'm trying to find the poles and determine the orders of the complex function $$f(z)=\frac{1}{(z+2i)^2}-\frac{z}{z-i}+\frac{z+1}{(z-i)^2}$$ I think that $$z=-2i$$ is a pole of order 2 and $$z=i$$ is also a pole of order 2, but I'm not sure how to show this. I know that if $$f(z)$$ has a pole at $$z=z_0$$ of order $$m$$ then the principal part of its Laurent series is finite: $$\frac{b_m}{(z-z_0)^m}+...+\frac{b_1}{z-z_0}$$ So, I tried to expand $$f(z)$$ in two Laurent series, one centered at $$z_0=-2i$$ and other at $$z_0=i$$.

For that I wrote $$-\frac{z}{z-i}+\frac{z+1}{(z-i)^2}=-1-\frac{i}{z-i}+\frac{z-i+i+1}{(z-i)^2}=-1-\frac{i}{z-i}+\frac{1}{z-i}+\frac{i+1}{(z-i)^2}$$ $$=-1+\frac{1-i}{z-i}+\frac{i+1}{(z-i)^2}$$

For the Laurent series of $$f(z)$$ centered at $$z=-2i$$, the first term is already part of this series, right? So I just need to express the sum of the second and third terms as the series centered at $$-2i$$. Those terms are analytic at $$-2i$$, right? So they can be expressed as a Taylor series only, so $$\frac{1}{(z+2i)^2}$$ would be the principal part of the Laurent series, making $$z=-2i$$ a pole of order 2.

For the Laurent series of $$f(z)$$ centered at $$z=i$$ the sum of the second and third terms are already part of this series. So I need to write the first term as a series centered at $$i$$, but this term is analytic at that point so it would be a Taylor series, making the sum of the last two terms the principal part of the Laurent series, making $$z=i$$ a pole of order 2.

Am I on the right path? How do I express the sum of the last two terms a series centered at $$z=-2i$$ and the first term as a series centered at $$z=i$$?

It is not necessary to find the exact Lauernt expansion: To find then order of pole at $$i$$, for example, you just observe that $$\frac 1 {(z+2i)^{2}}$$ is analytic in some disk around $$i$$ so it has a power series expansion of the type $$\sum\limits_{n=0}^{\infty} a_n (z-i)^{n}$$ in that disk. This gives the nature of the Laurent series expansion around $$i$$ and you can can infer that the oder of the pole is $$2$$.
• If the series expansion is of the type you mention, that would mean that the Laurent series is the sum of that series plus $-z/(z-i)+(z+1)/(z-i)^2$, right? And thus those two terms are the terms of the principal part in the Laurent series, making the pole of order 2. Is that what you meant? Commented Nov 28, 2020 at 5:50
• Yes, those two terms are the only ones with negative powers of $z-i$ so the order of the pole is $2$. @davidllerenav Commented Nov 28, 2020 at 5:52
• I see, the same would be for the pole at $-2i$. In this case, $-z/(z-i)+(z+1)/(z-i)^2$ is analytic in a disk around $z=-2i$ so it can be expanded in a power series with positive powers of $z+2i$ (its Taylor series), and thus $1/(z+2i)^{2}$ is the only term with negative powers of $z+2i$, so the order of the pole is once again 2 due to the power $-2$, right? Commented Nov 28, 2020 at 5:57