Discussion
Taken anticlockwise round the circle of radius two, the integral \( \oint dz/(z^{2}-1) \) comes to
Answer: A
Both poles \( z = \pm 1 \) lie inside \( |z| = 2 \), with residues \( 1/2 \) and \( -1/2 \), so
\[ \oint \frac{dz}{z^{2}-1} \;=\; 2\pi i\left(\tfrac{1}{2} - \tfrac{1}{2}\right) \;=\; 0. \]
The vanishing is a cancellation between the two residues, not a consequence of analyticity: a contour enclosing only one of them would give \( \pm\pi i \)
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