Question

Laurent series expansion of a function with removable singularities

score 262 · Answer 1 · 2018-08-25 13:49:46

262

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Laurent series expansion of a function with removable singularities

Published on 25 Aug 2018 - 13:49

score 851 · Answer 2 · 2018-08-29 16:07:28

851

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Show that the field of fractions of $\mathbb{Z}[[x]]$ is properly contained in $\Bbb Q((x))$

Published on 29 Aug 2018 - 16:07

score 93 · Answer 3 · 2018-08-29 21:24:08

93

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Laurent Expansion of $\frac{1}{f(z)^2}$, for $f$ a holomorphic function on a nbd of $a \in C$ with $f(a) = 0$ $f'(a) \ne 0$ and $f''(a) = 0$.

Published on 29 Aug 2018 - 21:24

score 57 · Answer 4 · 2018-09-08 12:34:20

57

Views

Integral of $e^\left(1/z^2\right)$ around $|z|=1$ in the complex plane

Published on 08 Sep 2018 - 12:34

score 39 · Answer 5 · 2026-03-25 10:55:50

39

Views

Poles of alternating series

Published on 25 Mar 2026 - 10:55

score 493 · Answer 6 · 2018-09-13 18:36:51

493

Views

Laurent series of $f(z)= \frac {z^2-2z+5}{(z-2)(z^2+1)}$

Published on 13 Sep 2018 - 18:36

score 77 · Answer 7 · 2018-09-15 23:00:36

77

Views

$\frac {1}{(x^2+1)^2}$ as a Laurent series

Published on 15 Sep 2018 - 23:00

score 360 · Answer 8 · 2026-03-25 09:29:23

360

Views

Principal part of Laurent Series for $\frac{1}{(1-\cosh(z))^2}$

Published on 25 Mar 2026 - 9:29

score 3.6k · Answer 9 · 2018-09-18 19:20:21

3.6k

Views

How can I find the residue of this removable singularity?

Published on 18 Sep 2018 - 19:20

score 643 · Answer 10 · 2018-09-19 15:04:27

643

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How can I evaluate this complex integral $\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz$?

Published on 19 Sep 2018 - 15:04

score 290 · Answer 11 · 2018-09-24 11:41:55

290

Views

Laurent Expansion of $f(z)=\frac{1}{z^2(1-z)}$

Published on 24 Sep 2018 - 11:41

score 39 · Answer 12 · 2018-09-24 20:28:06

39

Views

Help with understanding the contour integrals in Laurent series.

Published on 24 Sep 2018 - 20:28

score 57 · Answer 13 · 2018-09-25 00:57:25

57

Views

Laurent Series Question: Summation Starts from $-\infty$?

Published on 25 Sep 2018 - 0:57

score 78 · Answer 14 · 2018-09-25 06:51:32

78

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Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ that converges at 3

Published on 25 Sep 2018 - 6:51

score 6.1k · Answer 15 · 2018-09-25 09:02:27

6.1k

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Laurent series of $f(z)=\text{cosec}(z)$ that is valid at $\frac{1}{2}$

Published on 25 Sep 2018 - 9:02

List Question

Laurent series expansion of a function with removable singularities

Show that the field of fractions of $\mathbb{Z}[[x]]$ is properly contained in $\Bbb Q((x))$

Laurent Expansion of $\frac{1}{f(z)^2}$, for $f$ a holomorphic function on a nbd of $a \in C$ with $f(a) = 0$ $f'(a) \ne 0$ and $f''(a) = 0$.

Integral of $e^\left(1/z^2\right)$ around $|z|=1$ in the complex plane

Poles of alternating series

Laurent series of $f(z)= \frac {z^2-2z+5}{(z-2)(z^2+1)}$

$\frac {1}{(x^2+1)^2}$ as a Laurent series

Principal part of Laurent Series for $\frac{1}{(1-\cosh(z))^2}$

How can I find the residue of this removable singularity?

How can I evaluate this complex integral $\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz$?

Laurent Expansion of $f(z)=\frac{1}{z^2(1-z)}$

Help with understanding the contour integrals in Laurent series.

Laurent Series Question: Summation Starts from $-\infty$?

Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ that converges at 3

Laurent series of $f(z)=\text{cosec}(z)$ that is valid at $\frac{1}{2}$

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