Suppose \(f\) is holomorphic on \(\Omega\), a simply connected region, and suppose \(\gamma \subseteq \Omega\). Using the Cauchy integral formula, show that \[ \int_\gamma {f'(z) \over z-a}\dz = \int_\gamma {f(z) \over (z-a)^2 }\dz .\] Also prove this when \(\Omega\) is not simply connected.
Solution
Use the integral formula directly: \[ \int_\gamma {f'(z) \over z-a}\dz = 2\pi i f'(a) .\]
On the other hand, use Cauchy’s formula for derivatives: \[ \int_\gamma {f(z) \over (z-a)^2}\dz = 2\pi i f^{(1)}(a) ,\] and these values are equal.
If \(\Omega\) is not simply connected, note that by the quotient rule \[ \dd{}{z} {f(z) \over z-a} = {f'(z)\over z-a} - {f(z) \over (z-a)^2} .\]
Thus \[ \int_\gamma {f'(z) \over z-a} - \int_\gamma {f(z) \over (z-a)^2}\dz &= \int_\gamma \qty{ {f'(z) \over z-a} - {f(z) \over (z-a)^2} }\dz \\ &= \int_\gamma \dd{}{z} {f(z) \over z-a} \dz \\ &= G(\gamma(1)) - G(\gamma(0)) \\ &= G(p) - G(p) \\ &= 0 ,\] where \(G(z) \da {f(z) \over z-a}\) is a primitive for the integrand by definition.