Compute the following integrals: \[ \int_{\abs{z-1} = 1} {1\over z^2-1} \dz \\ \int_{\abs{z-0} = 2} {1\over z^2-1} \dz \\ .\] Compute the 2nd integral by finding a primitive.
Solution
For the first integral: \[ \int_\gamma{1\over z^2-1}\dz &= {1\over 2}\int_\gamma {1\over z-1} + {1\over z+1}\dz \\ &= {1\over 2}\int_\gamma {1\over z-1} \\ &= {1\over 2}\cdot 2\pi i \Res_{z=1}f(z) \\ &= {1\over 2}\cdot 2\pi i \cdot 1 \\ &= \pi i ,\] using that \(f(z) = {1\over z-1}\) is already an expansion of \(f\) about \(z=1\) since it is a Laurent series in \((z-1)^k\), so the residue is \(1\).
For the second integral: attempt to define a primitive \[ F(z) \da {1\over 2}\log\qty{z-1\over z+1} \implies F'(z) = {1\over z^2-1} .\] If this primitive is well-defined on \(\gamma\), the integral will vanish because this is a closed curve. Choose the branch cut \(\CC\sm(-\infty, 0]\) and define \(g(z) = {z-1\over z+1}\), so that \(F(z) = {1\over 2}\log(g(z))\). Then then \(g(z)\in (-\infty, 0] \iff z\in [-1, 1]\), and this is well-defined on \(\gamma\) since \([-1, 1]\) does not intersect \(\gamma\).
Thus for \(\gamma\) any parameterization of \(\abs{z} = 2\), \[ \int_{\abs{z} = 2} f(z) \dz = \int_{\abs{z} = 2} F'(z)\dz = F(\gamma(0)) - F(\gamma(1)) = 0 .\]