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I need to find all possible values of:

$$\int _C \frac{e^zdz}{z(z^2-1)}$$ when C is a circle that dosen't pass through 0,1,-1.

Can you please help?

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2 Answers 2

up vote 2 down vote accepted

Note that $f(z)=\dfrac{e^z}{z(z+1)(z-1)}$ has simple poles at $0,\pm 1$. Recall that at a simple pole $c$, the residue of $f$ is given by $$ \operatorname{Res}(f,c)=\lim_{z\to c}(z-c)f(z). $$ So in this case, one can calulate that $\operatorname{Res}(f,0)=-1$, $\operatorname{Res}(f,1)=\frac{e}{2}$, and $\operatorname{Res}(f,-1)=\frac{1}{2e}$.

There are $8$ possible subsets of $\{0,1,-1\}$ that such a circle $C$ as described in the problem may contain. Then use the fact that the value of the integral is $2\pi i$ times the sum of the residues of the poles $C$ encloses to find the possible values of the integral.

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Are you sure there are 8 possible values? Is there a circle that contain {-1,1} without containing {0}? –  Ran Kashtan Dec 29 '12 at 21:51
    
@tchernobilsky Oops, you're right. Let's skip that impossibility. –  Ben West Dec 30 '12 at 2:52

Hint: Use the Cauchy's Integral Formula or the Residue theorem as you see fit.

Here is an example: If $0,1,-1$ are not in the region bounded by the contour then by the Residue theorem (or the Integral theorem if you prefer), your integral is $0$.

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