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QUESTION


Use the Cauchy Residue Theorem to integrate each of the functions
around the circle $|z|=3$ in the counterclockwise sense. (a)
${e^{-z}\over z^2 }$, \ \ (b) ${e^{-z}\over (z-1)^2}$,\ \  (c)
$z^2e^{1\over z}$.



ANSWER


The integral is $2\pi i$ (sum of residues inside $|z|=3)$. In each
case there is only one pole inside $|z|=3$. (a) Pole at $z=0$;
  ${e^{-z}\over z^2}={1\over z^2}(1-z+{z^2\over 2!}+\cdots $. Thus residue at pole is -1
and so integral is $-2\pi i$.  (b) Pole at $z=1$. Put $z-1=w$.
 Then ${e^{-z}\over (z-1)^2}
={e^{-w+1}\over w^2}={e^{-w}\over w^2}e^{-1}$ and so by part (a)
the residue at $z=1$ is $-e^{-1}$ and so the integral is $-2\pi
ie^{-1}$. (c) $z^2e^{1\over z}=z^2(1+{1\over z}+{1\over
2z^2}+{1\over 6z^3}+\cdots)$ and so residue at $z=0$ is 1/6. Thus
integral is $2\pi i/6=\pi i/3$.




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