Question
Obtain the Laurent series expansion of about
. Determine the type of singularity and the region of convergence.
Answer :
Word Count : 57
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Let (w=z-1). Then [ \frac{e^{z}}{(z-1)^2}=\frac{e^{1+w}}{w^2}=e,\frac{e^{w}}{w^2}. ] Using the Taylor series (e^{w}=1+w+\frac{w^2}{2!}+\frac{w^3}{3!}+\cdots), we obtain [ \frac{e^{z}}{(z-1)^2} = e\left(\frac{1}{w^2}+\frac{1}{w}+\frac{1}{2!}+\frac{w}{3!}+\frac{w^2}{4!}+\cdots\right). ]
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