All Questions
148
questions
1
vote
1
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82
views
Find the Laurent expansion of $f(z) = \sin{\frac{1}{z(z-1)}}$ in $0<|z-1|<1$
Here is my idea:
$\sin{\frac{1}{z(z-1)}} = \sin{\left( \frac{-1}{z} + \frac{1}{z-1}\right)} = \sin{\left(\frac{1}{z-1} - \frac{1}{z}\right)} = \sin{\frac{1}{z-1}}\cos{\frac{1}{z}} - \cos{\frac{1}{z-1}}...
- 172
1
vote
1
answer
46
views
Taylor-Laurent series expansions
I'm having some issues finding how to series expand some complex functions that my professor gave past years in exams.
For example, in this exercise, it is asked to find the first two terms of the ...
- 113
2
votes
1
answer
45
views
Understanding the Laurent expansion of a meromorphic function about $\infty$.
Suppose $f:\hat{\mathbb{C}}\to\hat{\mathbb{C}}$ were meromorphic, and suppose $f$ has a pole at $\infty$. I'm trying to understand the Laurent series of $f$ about $\infty$. By definition, $f$ has a ...
- 730
2
votes
0
answers
34
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Laurent Series question for Exponentials
I must find the Laurent series for $f(z) = \frac{e^z}{z^2}$ in powers of $z$ for the annulus $ |z| > 0$.
I wrote $f(z) = \frac{1}{z^2} \sum_{n=0}^{\infty} \frac{z^n}{n!} = \sum_{n=0}^{\infty} \frac{...
- 247
1
vote
1
answer
28
views
Fast way to find the __regular__ part of the Laurent expansion of a function at a pole?
To evaluate a certain limit, I need to calculate the first term in the regular part of the Laurent expansion of the function
$$\frac{\pi}{\sin \frac{\pi(s+1)}{2}}$$
around -1 (should be the same thing ...
- 1,197
1
vote
2
answers
248
views
Prove that If singular part of Laurent series has infinite many terms, then $\lim_{z\to z_0}(z-z_0)^mf(z)$ doesn't exist for all nautural number $m$.
Given $f$ an analytic function in open $D \subset \mathbb C$, $z_0$ is an isolated singularity defined as $B(z_0;r)\backslash\{z_0\} \subset D$, then know that $f$ can be written as an expansion of ...
user1194243
1
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0
answers
29
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Find the Laurent series for the following function [duplicate]
Find the Laurent series for $f(z)=\frac{2z-3}{z^2-3z+2}$ centered in the origin and convergent in the point $z=\frac32$, specifying it's convergence domain.
So I'm having troubles understanding what ...
- 57
0
votes
2
answers
73
views
Solving Power Series Equations if we introduce Logarithmic Terms
If we have a complex power series equation like
$$
\sum_{n=0}^{\infty} a_n z^n = \sum_{n=0}^{\infty} b_n z^n
$$
then we can conclude $a_n = b_n$. We can see this by viewing $z^n$ as basis elements, or ...
- 351
0
votes
0
answers
41
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When is it necessary to expand out the first term(s) of a power series?
I'm comfortable with the process of finding the Laurent series for a complex function, but in many of the answers from the textbook the first few terms will be expanded from it. Since I'm teaching ...
- 381
1
vote
1
answer
92
views
Coefficients of Laurent series
I'm trying to find the coefficients $c_{-2}$, $c_{-1}$, $c_{0}$, $c_{1}$ and $c_{2}$ of the Laurent series around $z=0$ for the function $f(z)=\frac{e^z}{z(1-z)}$ in the region $1<|z|<\infty$. I'...
- 19
1
vote
2
answers
83
views
How to calculate the integral $\frac{1}{2\pi i}\oint_{|z|=1}\frac{2(1-\cos(z))e^z}{z^4}dz$
My try:
$$I = \frac{1}{2\pi i}\oint_{|z|=1}\frac{2(1-\cos(z))e^z}{z^4}dz$$
using the identity of Laurent's expansian $$a_{n} = \frac{1}{2\pi i}\oint_{C}\frac{f(z)}
{(z-z_{0})^{n+1}}dz.$$
I know for a ...
- 346
0
votes
1
answer
738
views
Power series expansion of 1/(1+z)^2 around z=1 [duplicate]
I'm struggling with getting the power series/Taylor series expansion of $f(z) = \frac{1}{(1+z)^2}$ around $z_0 = 1$.
Usually, I would do a partial fraction decomposition, and then do some re-arranging ...
- 9
1
vote
1
answer
74
views
Computing integral over unit circle
Compute \begin{equation}I=\frac{1}{2\pi i}\int_{C(0,1)}z^n\exp\left(\frac{2}{z}\right)\textrm{d}z\end{equation} where $C(0,1)$ is the unit circle centred at $0$ oriented anticlockwise, for integer ...
user1022331
0
votes
1
answer
54
views
Laurent Expansion of $\frac{(z+1)^2}{z(z^3+1)}$
I’m stuck on finding the Laurent expansion this function about $z=0$:
$$\frac{(z+1)^2}{z(z^3+1)}$$
What I tried was to compute the Binomial Expansion for the top bit and then expand:
$$\frac{1}{1+z^3}=...
- 531
0
votes
1
answer
27
views
Laurent expansion of function with singularities
I'm trying to get the power series expansion of the following function about $ z=0$:
$$f(z)=\frac{z^3}{1-z^4}$$
I'm having some troubles with this. Is it possible to do the following:
$$\frac{1}{1-z^4}...
- 531