Let \(\gamma\) be the closed curve in the complex plane that is given in polar coordinates by \(r = 2 + 3\cos\theta\), \(0 \leq \theta \leq 4\pi\), oriented in the direction of increasing \(\theta\). Let
\[f(z) = \frac{e^z}{2z - 1} + \frac{\sin(2z)}{(z - 2)^2} + \frac{\cos(5z)}{(z + 5i)^3}.\]
Calculate \(\int_\gamma f(z) \, dz\).
[Recall that in polar coordinates, \((-r, \theta)\) and \((r, \theta + \pi)\) give the same point in the plane.]
Solution
1. The winding numbers are determined by three quadratic polynomials.
Put \(P(w)=3w^2/2+2w+3/2\). For \(w=e^{i\theta}\), \[ P(w)=(2+3\cos\theta)e^{i\theta}=\gamma(\theta). \] The parameter \(w\) runs twice counterclockwise around \(|w|=1\). Consequently, for a point \(a\) off the curve, substitution in the index integral and the argument principle give \[ \operatorname{Ind}(\gamma,a) =2\frac1{2\pi i}\int_{|w|=1}\frac{P'(w)}{P(w)-a}\,dw =2N_a, \] where \(N_a\) is the number of zeros of \(P-a\) in \(|w|<1\), counted with multiplicity [@SS03].
For \(a=1/2\) the two roots are \((-2\pm i\sqrt2)/3\), of modulus \(\sqrt6/3<1\). For \(a=2\) the roots are \((-2\pm\sqrt7)/3\); the plus root has modulus less than one and the minus root has modulus greater than one, since \(2<\sqrt7<3\). Finally, on \(|w|=1\), \[ |3w^2/2+2w|\leq7/2<\sqrt{109}/2=|3/2+5i|. \] Rouché’s theorem compares \(P+5i\) with the nonzero constant \(3/2+5i\), giving no zeros in the disk [@SS03]. These calculations also exclude zeros on the unit circle for all three polynomials. Thus the curve avoids the poles and \[ \operatorname{Ind}(\gamma,1/2)=4,\qquad \operatorname{Ind}(\gamma,2)=2,\qquad \operatorname{Ind}(\gamma,-5i)=0. \]
2. The indexed residue formula evaluates the integral.
The only possible poles of \(f\) are \(1/2\), \(2\), and \(-5i\). At the first two, the residues are \[ \operatorname{Res}_{1/2}f=e^{1/2}/2, \qquad \operatorname{Res}_{2}f=(\sin(2z))'|_{z=2}=2\cos4. \] The other summands are holomorphic at each respective point. The third residue is multiplied by index zero, so contributes nothing. The residue theorem for a closed curve gives \[ \int_\gamma f(z)\,dz =2\pi i\left(4\frac{e^{1/2}}2+2\cdot2\cos4\right) =\boxed{4\pi i e^{1/2}+8\pi i\cos4} \] [@SS03]. The use of winding numbers accounts for both loops and both traversals, without assuming the curve simple.