# 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 $$z=i$$?

## 1 Answer

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? – davidllerenav Nov 28 '20 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 – Kavi Rama Murthy Nov 28 '20 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? – davidllerenav Nov 28 '20 at 5:57
• Yes, exactly! @davidllerenav – Kavi Rama Murthy Nov 28 '20 at 5:58
• Great, thank you so much! I was struggling to write the whole Laurent series so this helped a ton. – davidllerenav Nov 28 '20 at 6:01