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15
Math.TechQA.Club
2018-10-29 22:28:34
77
Views
Laurent Series for $f(z)=\frac{2}{(z+1)^2}-\frac{5}{z-5}$
Published on
29 Oct 2018 - 22:28
#complex-analysis
#proof-verification
#laurent-series
204
Views
Poles , Radius of Convergence and Laurent Series
Published on
30 Oct 2018 - 10:12
#complex-analysis
#laurent-series
323
Views
Radius of Convergence of Laurent Series
Published on
30 Oct 2018 - 10:51
#complex-analysis
#laurent-series
139
Views
Find the order of the pole of $f(z)=\frac{e^{-z}}{z(\cos{z}-1)}$ at $z=0.$
Published on
30 Oct 2018 - 16:58
#complex-analysis
#laurent-series
25
Views
Help calculating Residue at an essential singularity
Published on
01 Nov 2018 - 15:37
#complex-analysis
#laurent-series
1.4k
Views
Finding the Laurent series of $f(z) = e^{1/z}$ about $z_0 = 0$
Published on
02 Nov 2018 - 3:00
#complex-analysis
#laurent-series
64
Views
What is the Laurent series of $f(z) = \frac{1}{z^2 (z-3)^2}$ around $z = 3$
Published on
05 Nov 2018 - 20:57
#complex-analysis
#laurent-series
81
Views
Double checking a coefficient for a Laurent series, the series for $f(z) = \frac{z^2 + 1}{(z - i)^2}dz$
Published on
14 Nov 2018 - 4:37
#complex-analysis
#proof-verification
#laurent-series
644
Views
Laurent series advantage
Published on
14 Nov 2018 - 19:29
#power-series
#taylor-expansion
#laurent-series
38
Views
Laurent series $f(z)=\frac{\sqrt{z}}{z+i}$
Published on
15 Nov 2018 - 7:27
#complex-analysis
#proof-verification
#alternative-proof
#laurent-series
376
Views
General strategy in finding Laurent Series
Published on
15 Nov 2018 - 16:22
#complex-analysis
#laurent-series
260
Views
Calculate the Laurent series centered at i on an annulus
Published on
16 Nov 2018 - 0:23
#complex-analysis
#power-series
#taylor-expansion
#laurent-series
983
Views
Residue of order 3 -
Published on
17 Nov 2018 - 0:59
#complex-analysis
#residue-calculus
#laurent-series
63
Views
Verification of Laurent series for the function $f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4}$
Published on
17 Nov 2018 - 7:06
#complex-analysis
#proof-verification
#laurent-series
199
Views
Laurent series of $ \frac{z-12}{z^2 + z - 6}$ for $|z-1|>4$
Published on
19 Nov 2018 - 13:13
#complex-analysis
#laurent-series
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