Question

Why is the composition of convergent Laurent series convergent?

score 333 · Answer 1 · 2026-04-12 17:12:09

333

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Why is the composition of convergent Laurent series convergent?

Published on 12 Apr 2026 - 17:12

score 83 · Answer 2 · 2026-04-17 09:31:29

83

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Laurent series for $e^\frac{1}{z}$ about $z=0$

Published on 17 Apr 2026 - 9:31

score 285 · Answer 3 · 2026-05-10 22:58:31

285

Views

Show that the Laurent series expansion of $f(z)$ in the region $0 < |z-z_0| < r < R$ does not involve negative power of $(z-z_0)$.

Published on 10 May 2026 - 22:58

score 603 · Answer 4 · 2026-04-13 23:32:19

603

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Laurent series around z=-i for $\frac{1}{z(z-1)}$ and $1<|z+i|<\sqrt2$

Published on 13 Apr 2026 - 23:32

score 1.1k · Answer 5 · 2026-04-08 21:15:33

1.1k

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Finding Residues Using Partial Fractions

Published on 08 Apr 2026 - 21:15

score 145 · Answer 6 · 2026-04-13 10:50:02

145

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Laurent series for $f(z) = \frac{\sinh(z + 3i)}{z(z + 3i)^3}$ at $-3i$ (not leaving as a product of series)

Published on 13 Apr 2026 - 10:50

score 56 · Answer 7 · 2026-04-15 10:49:48

56

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An apparent contradiction based on considering the series of $f(z)$ and relating it to $f(1/z)$

Published on 15 Apr 2026 - 10:49

score 2k · Answer 8 · 2026-04-14 16:12:08

2k

Views

Find the two Laurent series for $f(z)=\frac{1}{z^3-z^4}$ that involve powers of z, and state their domains of convergence.

Published on 14 Apr 2026 - 16:12

score 404 · Answer 9 · 2026-04-17 05:17:34

404

Views

complex analysis Laurent series around no singularity point?

Published on 17 Apr 2026 - 5:17

score 162 · Answer 10 · 2026-05-10 16:32:49

162

Views

Explicit calculation of Laurent series coefficients by evaluating integral

Published on 10 May 2026 - 16:32

score 86 · Answer 11 · 2026-04-16 15:18:02

86

Views

Laurent Expansion Of $\frac{1}{z^2+1}$

Published on 16 Apr 2026 - 15:18

score 69 · Answer 12 · 2026-04-15 14:16:14

69

Views

Laurent Expansion $\frac{e^z}{z-1}$

Published on 15 Apr 2026 - 14:16

score 85 · Answer 13 · 2026-04-14 18:18:44

85

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Laurent series expansion of $f(z)=\frac 1{z^2+z+1}$ for $|z| > 1$ centered at $z=0$

Published on 14 Apr 2026 - 18:18

score 264 · Answer 14 · 2026-05-10 18:59:16

264

Views

Laurent Series for singularities and poles

Published on 10 May 2026 - 18:59

score 1.1k · Answer 15 · 2026-04-16 06:26:20

1.1k

Views

Laurent series $\frac{e^{2z}}{(z-1)^3}$ around $z=1$

Published on 16 Apr 2026 - 6:26

List Question

Why is the composition of convergent Laurent series convergent?

Laurent series for $e^\frac{1}{z}$ about $z=0$

Show that the Laurent series expansion of $f(z)$ in the region $0 < |z-z_0| < r < R$ does not involve negative power of $(z-z_0)$.

Laurent series around z=-i for $\frac{1}{z(z-1)}$ and $1<|z+i|<\sqrt2$

Finding Residues Using Partial Fractions

Laurent series for $f(z) = \frac{\sinh(z + 3i)}{z(z + 3i)^3}$ at $-3i$ (not leaving as a product of series)

An apparent contradiction based on considering the series of $f(z)$ and relating it to $f(1/z)$

Find the two Laurent series for $f(z)=\frac{1}{z^3-z^4}$ that involve powers of z, and state their domains of convergence.

complex analysis Laurent series around no singularity point?

Explicit calculation of Laurent series coefficients by evaluating integral

Laurent Expansion Of $\frac{1}{z^2+1}$

Laurent Expansion $\frac{e^z}{z-1}$

Laurent series expansion of $f(z)=\frac 1{z^2+z+1}$ for $|z| > 1$ centered at $z=0$

Laurent Series for singularities and poles

Laurent series $\frac{e^{2z}}{(z-1)^3}$ around $z=1$

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