# Questions tagged [laurent-series]

This tag is for questions about finding a Laurent series of functions and their convergence. The Laurent series is a generalisation of the power series which allows negative indices and is essential for investigating the behaviour of functions near poles.

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### Laurent series of $e^z$

find the Laurent series centered at $z=1$ $$f(z)=\frac{e^z}{(z-1)^2}$$ I thought that the denominator part is safe by our center and the expansion is just about the exponential which is a Taylor ...
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### Finding the Laurent series of $\frac{-z^2-2z+3i}{z-2i}$

Finding the Laurent series of $$f(z) = \frac{-z^2-2z+3i}{z-2i}$$ on the annulus of infinite external radius $|z-i| > 1$ This function looks fairly messy. We clearly have a simple pole at $z=2i$. ...
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### Laurent Series for $\frac{1}{1 - z}$ and $\frac{1}{(1 - z)^2}$ around $|z| > 1$.

I am supposed to find the Laurent series for $f(z) = \frac{1}{1 - z}$ and $f(z) = \frac{1}{(1 - z)^2}$ around $|z| > 1$. I think I can find the first one, but the second one to me seems a little ...
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### Why is $z\mapsto\frac{1}{z}$ the universal example of a holomorphic function that fails to have a primitive?

This Wikipedia article (on Morera's theorem) mentions the following in the introduction. In a certain sense, the $\frac{1}{z}$ counterexample is universal: for every analytic function that has no ...
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### Find Laurent expansion of $\frac{1}{\sin z^3}$ at $0 < ∣z-\sqrt{2 \pi}∣ < ∣\sqrt{4 \pi} - \sqrt{2 \pi} ∣$

I need to find the principal part of the Laurent expansion of the function in the annulus described in the title. In my attempt, I have replaced $z$ by $w:=z - \sqrt{2 \pi}$. I know the Taylor ...
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### Laurent series of $\frac{1}{(z^2 + 1)(z^2 - 9)}$

I want to derive Laurent series of $$f(z) = \frac{1}{(z^2 + 1)(z^2 - 9)}$$ for two sets: $1 < |z| < 3$ and $3 < |z|$. My work so far First thing to do is to apply partial fraction ...
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### Laurent Series and Residue for $f(z) = z^4 \sin(1/z)$ at $z=0$

I had to find the Laurent series and residue of the function. We know that the $res_{z_0} = a_{-1}$, the coefficient of $\frac{1}{z-z_0}$ of the Laurent series. For $f(z) = z^4 \sin(1/z)$ on the ...
How can I find the residue of $$\frac{\text{sin}(\frac{1}{z})}{z^2+1}\quad ?$$ at $z=0$? In order to find this, I tried to calculate $a_{-1}$ in the Laurent series, and then arrived at $$\frac{\text{... 0 votes 2 answers 40 views ### Calculate the integral \int_{\partial D} \frac{z+4}{z-4}\frac{e^z}{\sin z} dz. Take the rectangle D = \{ z\in\mathbb{C};|x|\leq 2,|y|\leq 1\}. I need to calculate the integral$$\int_{\partial D} \frac{z+4}{z-4}\frac{e^z}{\sin z} dz.$$The only singularity in this case is in ... 0 votes 1 answer 36 views ### Expand the Laurent Series for f(z) = \frac{1}{z(z-i)^2} for 0 < |z - i| < 1 My attempt As |z - i| < 1, this means, that$$\sum_{n=1}^{\infty} (z - i)^n = \frac{1}{1 - (z - i)}$$But more than that, I don't really know what to do. The problem here is, that what we get is: ... 1 vote 0 answers 41 views ### Find the Laurent Series for f(z) = \frac{1}{z(z-i)} for 1 < |z| < \infty My attempt We know, that$$\sum_{n=0}^{\infty} z^n = \frac{1}{1-z}$$for |z| < 1 As we have 1 < |z| < \infty, this means that we have such z that \left|\frac{1}{z}\right| < 1, thus ... 1 vote 1 answer 34 views ### Why is it that functions with nonisolated singularities at a point do not have Laurent series at that point? Learning complex analysis, I've been taught that a function like csc(1/z) cannot have a Laurent series at 0, because there is a nonisolated singularity there. If I recall correctly, one needs to not ... 1 vote 1 answer 78 views ### What is the Laurent series of f(z)=1/z^2 I have just looked at about 10 different Laurent series in the past days, but I cannot solve f(z)=\frac{1}{z^2}. I attempt, |z|>0, with u=z^2+1: \begin{equation} \frac{1}{z^2}=\frac{1}{z^2-1+... 0 votes 1 answer 25 views ### What's with the lower bound of Laurent series? I have the simple example of finding the Laurent series of: f(z)=\frac{1}{z-2} on |z|\leq2 and |z|\geq2 Procedure to find it is rather straight forward, develop it as a Power series (in the form ... 0 votes 1 answer 31 views ### Developing Laurent series on three types of regions: centered at the origin, and non-centered at the origin So, in lieu of this post, I want to deepen the discussion of the Laurent series. We have the Laurent theorem: If f(z) is analytic on R_1\leq |z-z_o|\leq 2, then f(z)=\sum_{n=-\infty}^\infty c_n(z-... 2 votes 1 answer 44 views ### How to grasp the relationship between Laurent series depending on the region they are developed at in trying to understand how to set up the Laurent series for a fractional expression, I have the given function \begin{equation} \frac{1}{(z-2)}-\frac{1}{(z-1)} \end{equation} The Laurent series of ... 0 votes 1 answer 22 views ### Finding a Laurent series for a complex function with two poles where one is outside of the region when finding the Laurent series for \frac{1}{(z-1)(z-2)} in the region |z| <1, I thought that one should first evaluate where the poles are, inside or outside the given region. z=2 is outside ... 0 votes 1 answer 60 views ### Laurent Series of f(z) = \frac{(z+1)^2}{z(z^3+1)} about z = 0? This is supposed to be a non-calculator question, so I managed to get this far;  z^3 + 1 = (z + 1)(z^2 - 1 +1) , by polynomial division. Therefore,  f(z) = \frac{z+1}{z(z^2-z+1)} = (1 + \frac{1}{... -2 votes 1 answer 45 views ### Challenging Laurent series I want to find the Laurent series expansion of f(z)=\frac{e^{2z}}{z^2}, and apprently, the pole is at 0, so this series would be for |z|>0. I am no sure this is correct, but if I consider the ... 2 votes 1 answer 84 views ### Manipulation of Infinite Series Example I'm struggling to understand the manipulation of an infinite series shown in the text below. We begin with the series \sum_{n=0}^{\infty} u^n which we know converges since |u|<1. We then ... 0 votes 1 answer 40 views ### Finding the Laurent series when a residue at z=0 is given I have a_n=\frac{1}{n^4+1}, where n is an integer. I want to determine the Laurent series for a function f(z) such that the residue for \frac{f(z)}{z^n} in z=0 is a_n \forall \ n. Based on ... 0 votes 0 answers 33 views ### Meromorphic function can be represented by a rational function Let z_1,...,z_n\in \mathbb{C} and f: \mathbb{C}\setminus \{z_1,...,z_n\} \to \mathbb{C} be holomorphic. Furthermore, \operatorname{ord}(f,z_i) is finite for all i and \lim\limits_{|z|\to\... 0 votes 1 answer 57 views ### Laurent series outside region of convergence of \frac{1}{z^2+1} I have already computed the Laurent series of \frac{1}{z^2+1} at z_0=i in R=\{z\in \mathbb{C} : 0<|z-i|<2\}. I have to compute the Laurent series in |z-i|>2 right now, how can I do ... 1 vote 1 answer 50 views ### How does one arrive at the basic limit formulation for the Stieltjes constants? The starting definition that I am using is:$$\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^\infty\gamma_n\cdot\frac{(-1)^n}{n!}(s-1)$$If I naively differentiate, I find:$$\zeta'(s)=\sum_{m=1}^\infty-\frac{\ln(...
How can I find the order of the pole $z = \frac{\pi}{2}$ for $f(z)=\frac{1}{(2\log(z))(1-\sin(z))}$? I know the answer should be 2, but I can't solve it, mostly due to poor understanding of the pole ...