Answer to Question #102647 in Complex Analysis for BIVEK SAH

Question #102647
Obtain the Laurent series expansion of the function
e^z/(z-2)^2
about the singularity
z =2
1
Expert's answer
2020-02-20T08:26:54-0500

Obtain the Laurent series expansion of the function

e z/(z-2)2

about the singularity

z =2

Solution:

"{e^z\\above{2pt}(z-2)^2}=e^2 {e^z e^{-2}\\above{2pt}(z-2)^2 }".

Recall that

e z="\\displaystyle\\sum_{n=0}^\u221e" "{z^n\\above{2pt} n!}" for |z|<∞.

Using this we obtain the Laurent series expansion for "\\left(e^{\\smash{z-2}}\\right)=\\displaystyle\\sum_{n=0}^\u221e {(z-2)^n\\above{2pt} n!}"

for |z-2|<∞.Thus,

"{e^z\\above{2pt} (z-2)^2}=e^2 {1 \\above{2pt} (z-2)^2} \\displaystyle\\sum_{n=0}^\u221e {(z-2)^n \\above{2pt} n!}"

= "e^2\\displaystyle\\sum_{n=0}^\u221e {(z-2)^n\\above{2pt} n!}" "{\\left(z-2^{\\smash{}}\\right)^{-2} \\above{2pt} n!}"

="e^2 [{1 \\above{2pt} (z-2)^2}+{1\\above{2pt} z-2}+\\displaystyle\\sum_{n=2}^\u221e { (z-2)^{n-2}\\above{2pt} n!}]"

= "e^2 [\\displaystyle\\sum_{n=0}^\u221e {(z-2)^n \\above{2pt} (n+2)!}+{1 \\above{2pt} z-2}+{1 \\above{2pt} (z-2)^2}]"




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