Question

Is this the correct way to find the Laurent Expansion of the complex function $f(z) = \frac{z}{z^2 - 1}$?

score 55 · Answer 1 · 2017-05-02 17:14:41

55

Views

Is this the correct way to find the Laurent Expansion of the complex function $f(z) = \frac{z}{z^2 - 1}$?

Published on 02 May 2017 - 17:14

score 64 · Answer 2 · 2017-05-03 05:13:58

64

Views

What happens to the isolated singular points of functions f and g when you multiply the functions?

Published on 03 May 2017 - 5:13

score 248 · Answer 3 · 2017-05-03 11:53:21

248

Views

Find and classify singular points

Published on 03 May 2017 - 11:53

score 61 · Answer 4 · 2017-05-03 13:49:05

61

Views

If a function $f$ is not analytic at a point $a$, can we conclude that $f^2$ will not be analytic at $a$?

Published on 03 May 2017 - 13:49

score 121 · Answer 5 · 2026-03-29 15:23:21

121

Views

Does this analytic function is zero?

Published on 29 Mar 2026 - 15:23

score 461 · Answer 6 · 2026-03-28 13:22:10

461

Views

How can I classify the singularities of a function, without calculating the Laurent series?

Published on 28 Mar 2026 - 13:22

score 282 · Answer 7 · 2017-05-06 15:10:32

282

Views

Laurent Series of $f(z)=\sin(\frac{1}{z})$ about $z=0$

Published on 06 May 2017 - 15:10

score 250 · Answer 8 · 2017-05-07 11:22:22

250

Views

Finding Laurent Series-is it possible without contour integral

Published on 07 May 2017 - 11:22

score 202 · Answer 9 · 2017-05-07 21:25:59

202

Views

Laurent expansion in annular region

Published on 07 May 2017 - 21:25

score 146 · Answer 10 · 2017-05-10 15:44:16

146

Views

Identity involving $\log$ and a power series

Published on 10 May 2017 - 15:44

score 164 · Answer 11 · 2017-05-12 13:05:53

164

Views

Laurent series of $ \frac{z-12}{z^2 + z - 6}$ valid for $|z-1| >1$

Published on 12 May 2017 - 13:05

score 234 · Answer 12 · 2017-05-12 18:36:09

234

Views

Laurent series of $e^{\frac{1}{z-1}}$ for $|z|>1$

Published on 12 May 2017 - 18:36

score 625 · Answer 13 · 2026-03-25 16:03:02

625

Views

Integral: $\int_{|z|\mathop=2}\sqrt{z^4-z}\,dz$ - finding Residues / Laurent Series

Published on 25 Mar 2026 - 16:03

score 2.4k · Answer 14 · 2017-05-14 10:59:25

2.4k

Views

Find the Laurent series of $1/(z^3+1)$ in the annulus $1<|z|<3$

Published on 14 May 2017 - 10:59

score 100 · Answer 15 · 2017-05-14 12:52:20

100

Views

Laurent Series for $f(z)=\frac{9z-1}{(z^2-9)(z^3+1)}$ in annulus $1<|z|<3$

Published on 14 May 2017 - 12:52

List Question

Is this the correct way to find the Laurent Expansion of the complex function $f(z) = \frac{z}{z^2 - 1}$?

What happens to the isolated singular points of functions f and g when you multiply the functions?

Find and classify singular points

If a function $f$ is not analytic at a point $a$, can we conclude that $f^2$ will not be analytic at $a$?

Does this analytic function is zero?

How can I classify the singularities of a function, without calculating the Laurent series?

Laurent Series of $f(z)=\sin(\frac{1}{z})$ about $z=0$

Finding Laurent Series-is it possible without contour integral

Laurent expansion in annular region

Identity involving $\log$ and a power series

Laurent series of $ \frac{z-12}{z^2 + z - 6}$ valid for $|z-1| >1$

Laurent series of $e^{\frac{1}{z-1}}$ for $|z|>1$

Integral: $\int_{|z|\mathop=2}\sqrt{z^4-z}\,dz$ - finding Residues / Laurent Series

Find the Laurent series of $1/(z^3+1)$ in the annulus $1<|z|<3$

Laurent Series for $f(z)=\frac{9z-1}{(z^2-9)(z^3+1)}$ in annulus $1<|z|<3$

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