Without using Cauchy’s integral formula, show that if \(\abs{a} < r < \abs{b}\), then \[ \int_{\gamma} \frac{d z}{(z-a)(z-b)} =\frac{2 \pi i}{a-b} \] where \(\gamma\) denotes the circle centered at the origin of radius \(r\) with positive orientation.
Hint: take a Laurent expansion.
Solution
Since \(|a|<r<|b|\), on \(|z|=r\) we may expand \[ {1\over z-a} ={1\over z}{1\over1-a/z} =\sum_{m=0}^\infty a^m z^{-m-1} \] and \[ {1\over z-b} =-{1\over b}{1\over1-z/b} =-\sum_{n=0}^\infty {z^n\over b^{n+1}}. \] Both series converge uniformly on the circle. Multiplying them gives a Laurent series for the integrand. Its coefficient of \(z^{-1}\) is \[ -\sum_{m=0}^\infty {a^m\over b^{m+1}} =-{1\over b-a} ={1\over a-b}. \] Termwise integration around \(\gamma\) therefore gives \[ \int_\gamma {dz\over(z-a)(z-b)} =2\pi i\,{1\over a-b}, \] because every Laurent monomial except \(z^{-1}\) integrates to zero.