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How I can solve this complex integral?

$$\int_{|z|=2}\dfrac{e^z}{z(z^2-1)(z+1)}dz$$

I know how to find the singularities but I do not understand in this case how to distribute the fractions of the integral. I mean $A/(x-a)$ etc.... The singularities are: $z=0$,$z=\pm 1$, $z=-1$. I got stuck.

Thank you in advance for your help.

Robert Z
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user495372
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1 Answers1

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With Residue theorem we see all denominator zeros $z=0,1,-1$ lie in circle $|z|=2$, for $f(z)=\dfrac{e^z}{z(z^2-1)(z+1)}$ \begin{align} \operatorname{Res}_{z=0}f(z)=\lim_{z\to0}(z-0)f(z)=-1,\\ \operatorname{Res}_{z=1}f(z)=\lim_{z\to1}(z-1)f(z)=\dfrac{e}{4},\\ \operatorname{Res}_{z=-1}f(z)=\lim_{z\to-1}[(z+1)^2f(z)]'=\dfrac{5}{4e},\\ \end{align} then $$\int_{|z|=2}\dfrac{e^z}{z(z^2-1)(z+1)}=2\pi i\sum_{0,1,-1} \operatorname{Res}=2\pi i\left(-1+\dfrac{e}{4}+\dfrac{5}{4e}\right)$$

Nosrati
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  • OP says "I do not understand in this case how to distribute the fractions of the integral". Does OP know Residue theorem? – Robert Z Nov 21 '17 at 15:54
  • Thank you very much for the response, I forgot to try with the residue method-theorem. Thanks for your help! – user495372 Nov 21 '17 at 16:03