Answer to Question #102647 in Complex Analysis for BIVEK SAH

Question

Obtain the Laurent series expansion of the function
e^z/(z-2)^2
about the singularity
z =2

Answer 1

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

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

for |z-2|<∞.Thus,

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

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

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

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