Let \[ f(z)=z^{-2}(1-z^2)^{-1}e^z. \] Evaluate \[ \frac1{2\pi i}\int_\gamma f(z)\,dz, \] where \(\gamma\) is the curve depicted in Problem 12 of the Berkeley Fall 1993 source exam.
Solution
1. The function\[ f(z)=\frac{e^z}{z^2(1-z^2)} \]has singularities only at \(0\), \(1\), and \(-1\), with\[ \Res_{z=0}f(z)=1, \qquad \Res_{z=1}f(z)=-\frac e2, \qquad \Res_{z=-1}f(z)=\frac{e^{-1}}2. \]
At \(0\) there is a double pole. Writing \[ f(z)=\frac1{z^2}\frac{e^z}{1-z^2}, \] its residue is \[ \left.\frac{d}{dz}\left(\frac{e^z}{1-z^2}\right)\right|_{z=0} =1. \]
The poles at \(1\) and \(-1\) are simple. Since \[ 1-z^2=-(z-1)(z+1), \] one obtains \[ \Res_{z=1}f(z) = \lim_{z\to1} \frac{-e^z}{z^2(z+1)} = -\frac e2. \] Similarly, using \(1-z^2=(1-z)(z+1)\), \[ \Res_{z=-1}f(z) = \lim_{z\to-1} \frac{e^z}{z^2(1-z)} = \frac{e^{-1}}2. \]
2. If \(\gamma\) is closed, piecewise smooth, and avoids\(0,1,-1\), then\[ \frac1{2\pi i}\int_\gamma f(z)\,dz = \boxed{ \operatorname{Ind}_\gamma(0) -\frac e2\operatorname{Ind}_\gamma(1) +\frac{e^{-1}}2\operatorname{Ind}_\gamma(-1) }. \]
The residue theorem with winding numbers gives \[ \frac1{2\pi i}\int_\gamma f(z)\,dz = \sum_{a\in\{0,1,-1\}} \operatorname{Ind}_\gamma(a)\Res_{z=a}f(z). \] Substituting the three residues from step 1 gives the displayed formula.
3. Q.E.D.
Step 2 gives the integral in terms of the winding numbers \(\operatorname{Ind}_\gamma(0)\), \(\operatorname{Ind}_\gamma(1)\), and \(\operatorname{Ind}_\gamma(-1)\) of \(\gamma\) about the three poles of \(f\).