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Problem 11.4.2 Arfken

Evaluate $\oint_C \frac{dz}{z^2-1}$, where $C$ is the circle $|z-1||=1$.

My approach to the solution is that to express the integral as partial fractions.

$\oint_C \frac{dz}{z^2-1}=\frac12 \oint_C ( \frac1{z-1} - \frac1{z+1} ) dz$. Since the point $z=1$ is within the contour, and $z=-1$ is outside the contour, using Cauchy's Integral Formula, one gets $\pi i$. But the answer in the instructor's manual is $0$. So where am I going wrong?

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  • $\begingroup$ The answer given here is also $i\pi$. $\endgroup$ – Rohan Dec 23 '16 at 5:21
  • $\begingroup$ using Cauchy integral's formula and Cauchy integral's theorem. Otherwise you can write $\frac{1}{z^2-1} = \frac{1/(z+1)}{z-1}$ And your result is correct. $\endgroup$ – reuns Dec 23 '16 at 5:37
  • $\begingroup$ Thanks. The solution manual needs a revision then. Residues too gives me $\pi i$. $\endgroup$ – sbp Dec 23 '16 at 6:35

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