Laurent series of $f(z)=\dfrac{4z-z^2}{(z^2-4)(z+1)}$ in different annulus

Given $ f(z)=\dfrac{4z-z^2}{(z^2-4)(z+1)}$ I need to find the Laurent series in the annulus: $ A_{1,2}(0),\;A_{2,\infty}(0),\;A_{0,1}(-1)$

I found the following partial fractions: $ f(z)=\dfrac{-3}{(z+2)}+\dfrac{1}{3(x-2)}+\dfrac{5}{3(x+3)}$ ,

the power series of these fractions are:

$ \dfrac{-3}{(z+2)}=\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{-z}{2} \right)^n} $

$ \dfrac{1}{3(z-2)}=\displaystyle{\frac{-1}{6}\sum_{n=0}^\infty \left( \frac{-z}{2} \right)^n} $

$ \dfrac{5}{3(z+1)}=\displaystyle{\frac{5}{3}\sum_{n=0}^\infty \left( -z \right)^n} $

and the principle parts are:

$ \dfrac{-3}{(z+2)}=\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{1}{-2z} \right)^n} $

$ \dfrac{1}{3(z-2)}=\displaystyle{\frac{-1}{6}\sum_{n=0}^\infty \left( \frac{1}{-2z} \right)^n} $

$ \dfrac{5}{3(z+1)}=\displaystyle{\frac{5}{3}\sum_{n=0}^\infty \left( \frac{1}{-z} \right)^n} $

In the first annuli I take the principle part only of $ \dfrac{5}{3(z+1)}$ , in the second annuli I take the principle part of all fraction. About the third one, I have $ 0<\vert z-1\vert<1$ , I denoted $ w=z-1$ and then I took the power series for all fractions and simply switched the $ w$ back to $ z-1$ at the end. Is it the right way of doing it?

I recieved $ \displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{1-z}{2} \right)^n – \frac{1}{6}\sum_{n=0}^\infty \left( \frac{1-z}{2} \right)^n + \frac{5}{3}\sum_{n=0}^\infty \left( 1-z \right)^n}$

Is this a correct Laurent series expansion for the given annulus?

Expand the function $ $ f(z) = \frac{1}{(z + 1)(z + 3)}$ $ in a Laurent series valid for $ 1 < |z| < 3$


My attempt:

$ $ \frac{1}{(z + 1)(z + 3)}=\frac{1}{4}.\frac{1}{1+z}-\frac{1}{4}.\frac{1}{3+z}$ $

$ $ =\frac{1}{4}.\frac{1}{1-(-z)}-\frac{1}{4}.\frac{1}{3-(-z)}$ $

$ $ =\frac{1}{4}.\frac{1}{z}\frac{1}{\frac{1}{z}-(-1)}-\frac{1}{4}.\frac{1}{3-(-z)}$ $

$ $ =\frac{1}{4}.\frac{1}{z}\frac{1}{1-(-\frac{1}{z})}-\frac{1}{4}.\frac{1}{3}.\frac{1}{1-(-\frac{z}{3})}$ $

and now both fraction can be expanded using the geometric series

$ $ =\frac{1}{4}.\frac{1}{z}(1-\frac{1}{z}+\frac{1}{z^2}…)-\frac{1}{12}.(1-\frac{z}{3}+\frac{z^2}{3^2}…)$ $


Is the expansion correct?

NOTE: What is written above is the entire question.

Integration of complex variables using Laurent series about a contour, C

I’ve got a homework question that I believe requires me to use Laurent series/method of residues.

The question itself is: Evaluate $ \int_C \frac{1}{z^2(z^2-16)}$ where C is the contour $ |z| = 1$ . I’m confused by this question because it doesn’t say anything about the orientation of C. I’m ashamed to say I don’t even know how to approach this.

So far, I’ve tried breaking it into: $ \int_C \frac{1}{z^2} \frac{1}{z+4i} \frac{1}{z-4i}$ . But I don’t really know how to get the Taylor/Laurent series for these three pieces.

Am I approaching this in the right way?
Can somebody help me move forward?

Oops! I just realized that this is the wrong place to ask! Sorry!

Fourier coefficients of $f\left(z\right)=\frac{1}{1+\cos^{2}z}$ through Laurent series

I had to find the Fourier coefficients of this simply periodic function $ $ f\left(z\right)=\frac{1}{1+\cos^{2}z},$ $ I proceeded considering the $ w=exp(iz)$ and considering the Laurent expansion of the function $ F(w)$ $ $ F\left(w\right)=\frac{1}{1+\frac{1}{2w}\left(w^{2}+1\right)}=\frac{2w}{\left(w+1\right)^{2}},$ $ If I’m not making a mistake, this function has a Leurent expansion in zero given by $ $ F\left(w\right)=\sum\left(-1\right)^{n}2\left(n+1\right)w^{n+1},\,\,\,n\geq0$ $ So writing back fourier expansion for $ f(z)$ I have the following Fourier expansion $ $ f\left(z\right)=\sum c_{n}e^{inz},\,\,\,\,\,c_{n}=\left(-1\right)^{n}2n,\,\,\,\,n\geq1$ $ I think everything is fine but when I’m doing the same analysis with an equivalent formulation of $ f(z)$ I have a different result. Can anybody check for me if those I found are the real Fourier coefficients of my $ f(z)$ ?

On the Newton polygon for Laurent series

I’m stuck with an understanding of what should be the Newton polygon for a Laurent series. I’m reading ”An introduction to G-function” by Dwork and he dedicates only three pages to Newton polygons for Laurent series; he says what is a Laurent series, how we may ”decompose” it as the sum of two power series, and only deals the case when we have an annulus of convergence. The purpose of that section is to give a sketch of the proof of the generalized Weierstrass preparation theorem (A version of that theorem but for Laurent series). I’m filling the gaps, completing things and adding information. I have everything, even the theorem which in some sense the proof is similar to the case of power series; but what I dont understand is how the Newton polygon should be. Here is my attempt:

Recall that a Laurent series, is a series of the form $ $ f(X)=\sum_{n=-\infty}^{\infty}a_{n}X^{n}, \quad a_{n}\in\mathbb{Q}_{p}$ $

($ v$ is the additive valuation on $ \mathbb{Q}_{p}$ )

We can write $ f(X)=f^{+}(X)+f^{-}(1/X)$ where $ $ f^{+}(X)=\sum_{n=0}^{\infty}a_{n}X^{n}\quad\text{and}\quad f^{-}(X)=\sum_{n=1}^{\infty}a_{-n}X^{n}\text{.}$ $

We define the Newton polygon of $ f$ as in the case of power series. This is his ”definition”. So I tried to understand what it means:

So, the Newton polygon of $ f$ is the convex hull of the set of point $ (j,v(a_{j}))$ . In power series if $ a_{j}=0$ we think the point $ (j,v(a_{j}))$ as the point at infinity on the upper half plane but now the problem is consider the points when $ n<0$ . Lets take this examples:

  1. Consider $ \sum_{n\in\mathbb{Z}}pX^{n}$ , we have $ (j,v(a_{j}))=(j,0)$ for every $ n\in\mathbb{Z}$ , so, I infer that the Newton polygon must be all the x-axis.

Recall that in power series the set of points are on or above the Newton polygon, but how this condition translates into Laurent series? this question comes from the next example:

  1. Consider $ f=\sum_{n=-\infty}^{\infty}p^{n}X^{2^{n}}$ . Here its more complicated. I tried to obtain the Newton polygon of $ f$ from the Newton polygons of $ f^{+}$ and $ f^{-}$ . The Newton polygon of $ f^{+}$ is the positive $ x-$ axis. But what happens with the Newton polygon of $ f^{-}$ ? The points $ (j,v(a_{j}))$ with $ j<0$ are on the negative $ x-$ axis and if $ n$ is not a power of $ 2$ the point $ (j,v(a_{j}))$ seems to be the point at minus infinity. I think that the Newton polygon of $ f^{-}$ must be the negative part of the $ x-$ axis so that the Newton polygon of $ f$ must be all the $ x-$ axis.

I’m starting with known power series. These examples comes from my examples on power series, and the Newton polygon of these have a finite number of sides. Also, in power series I start with an infinite vertical line, then one constructs the Newton polygon, but in Laurent series I don’t have any vertical line (we only deal with convergent Laurent series). But what happens if we start from a power series having an infinite number of sides and trying to expand over $ n\in\mathbb{Z}$ ? For example, the Newton polygon of the logarithm is well known, and looking at what ”should” be the Newton polygon of $ f^{-}$ it makes me to think the following:

I can define the Newton polygon of a Laurent series $ f$ by the following: Let $ C(f^{+})$ the convex hull of the set of points $ (j,v(a_{j}))$ where $ j\geq 0$ and let $ C(f^{-})$ the convex hull of the set of points $ (j,v(a_{j}))$ with $ j<0$ . Then the Newton polygon of $ f$ is the convex hull of $ C(f^{+})\cup C(f^{-})$ . If $ a_{j}=0$ for $ j\geq 0$ we put $ (j,v(a_{j}))=+\infty$ and if $ a_{j}=0$ for $ j<0$ we put $ (j,v(a_{j}))=-\infty$ . If $ n<0$ the set of points $ (j,v(a_{j}))$ are on or below $ C(f^{-})$ . Explicitly for $ C(f^{-})$ we start from the positive $ y-axis$ and we turn it clockwise and we follow the rules of the process with $ f^{-}$ but in this direction, so I think that $ C(f^{-}(X))=C(f^{+}(-X))$ .

Is this ”definition” (conjecture) valid? I was looking for books having this material, but I only found “A course in $ p$ -Adic Analysis” by Alain Robert. This is the only reference I have in English. In French I have ”Les nombres p-adiques” by Yvette Amice. Here defines the Newton polygon of Laurent series as the graph of certain function and then proves that is equivalent to: The Newton polygon of $ f$ is the boundary of the convex upper envelope of the set of points $ (j,v(a_{j}))$ . Then there are some examples of Newton polygons, but the examples are polynomials or power series, she never considers Laurent series.

I really appreciate any reference, discussion, if my ”definition” is valid, anything. As I said, I’m filling the gaps and other stuffs, but also I want to have examples of what I’m saying.

Thanks a lot, also forgive me if I have misprints, I’m from Mexico and I’m not an expert on English. Greetings

“Lambert Polynomials” vs Laurent Polynomials

I was reading this post on why polynomials can’t have negative exponenents.
The most voted answer seems to bring out a difference between some objects called “Lambert Polynomials” and Laurent Polynomials.
These “Lambert polynomials” are cited as a counterexample to rational functions becasue they miss the property of being closed under division.
As reported in the original post “This property doesn’t hold for your ‘Lambert polynomials’, because there’s no finite expression in positive and/or negative powers of x that corresponds to the function $ \displaystyle \frac{1}{1+x}$ .”

Then the author conludes explaining the notion of a Laurent Polynomial.
I didn’t get the differenct between the two notions as it was laid out in that post, but I’m very interested in the topic.
I hope someone can clarify that for me.

Calculation of Laurent series and decomposition in partial fractions

So I was asked to find the Laurent series of the expression around z=2: $ $ f(z)= \frac{z+3}{(z-2)^3}$ $

I am aware that this can be decomposed as $ $ 5\, \left( z-2 \right) ^{-3}+ \left( z-2 \right) ^{-2}$ $ Is this the entire Laurent series of $ f(z)$ around $ z=2$ . I am asking because in order to derive this I haven’t even used the fact that expansion is being carried out about 2. If so why?

Advice on using this alternative method of finding the Laurent expansion of $\tfrac{1}{z^2(z-1)}$

Say we want to calculate the Laurent series of $ \tfrac{1}{z^2(z-1)}$ about $ z_0=1.$ Now I know that one way to do it is to say that $ f(z)=\tfrac{1}{z^2}(\tfrac{1}{z-1})$ and appy the geometric series expansion to the brackets term. But I wanted to try and do it a different way :

First we split f into partial fractions and compute the Laurent series separately.Now consider the Laurent expansion of $ \tfrac{1}{z^2}$

We know that $ 0$ is a pole of order 2 which implies that $ \forall n>2,a_{-n}=0$ . Therefore $ \tfrac{1}{z^2}$ haa series expansion $ \tfrac{a_{-2}}{(z-z_0)^2}+\tfrac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+…$

Now to calculate the the $ a_{-2}$ coefficient I applied the following trick

$ a_n=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{f(z)}{(z-z_0)^{n+1} }dz=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{(z-z_0)^{n+3}}dx=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{z^{n+3}}dz$

$ a_{n-2}=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{(z-z_0)^{n+1} }dz=\tfrac{f^n(z_0)}{n!}$ But as n $ f(z)=1$ this implies that $ n$ must be zero and so $ a_{-2}=1$

Now when I tried to use the same trick on $ a_{-1}$ It doesn’t work because now we can’t use Cauchy’s formula. Also when I tried to use u substitution by letting $ u=z-z_0$ it returns that $ a_{-1}=-\tfrac{1}{z}$ but I know this is not rue as I know from the method that I mentioned in the first paragraph that $ a_{-1}=1$ So does anyone have any suggestions on how I can find $ a_{-1}$ continuing with the method I’m trying to use ?