Let \(n\in \ZZ^{\geq 1}\) and \(0<\theta<\pi\) and show that \[ {1\over 2\pi i} \int_{\abs z = 2} {z^n \over 1 -3z\cos(\theta) + z^2} \,dz = {\sin(n\theta) \over \sin(\theta)} .\]
Solution
1. Factorization of the quadratic denominator \(P(z) = z^2 - 2z\cos\theta + 1\):
1.1. For \(0 < \theta < \pi\), the quadratic \(z^2 - 2z\cos\theta + 1\) factors over \(\mathbb{C}\) as:\[ z^2 - 2z\cos\theta + 1 = (z - e^{i\theta})(z - e^{-i\theta}). \]
1.2. The roots are \(z_1 = e^{i\theta}\) and \(z_2 = e^{-i\theta}\).Since \(|z_1| = |z_2| = 1 < 2\), both poles lie strictly inside the circular contour \(\gamma: |z| = 2\).Since \(0 < \theta < \pi\), \(\sin\theta \neq 0\), so \(z_1 \neq z_2\), and both poles are simple poles.
2. Residue computation:
2.1. Let \(g(z) = \frac{z^n}{(z - e^{i\theta})(z - e^{-i\theta})}\).At the simple pole \(z_1 = e^{i\theta}\):\[ \operatorname{Res}(g, e^{i\theta}) = \lim_{z \to e^{i\theta}} (z - e^{i\theta}) g(z) = \frac{(e^{i\theta})^n}{e^{i\theta} - e^{-i\theta}} = \frac{e^{in\theta}}{2i\sin\theta}. \]
2.2. At the simple pole \(z_2 = e^{-i\theta}\):\[ \operatorname{Res}(g, e^{-i\theta}) = \lim_{z \to e^{-i\theta}} (z - e^{-i\theta}) g(z) = \frac{(e^{-i\theta})^n}{e^{-i\theta} - e^{i\theta}} = -\frac{e^{-in\theta}}{2i\sin\theta}. \]
2.3. Summing the residues:\[ \operatorname{Res}(g, e^{i\theta}) + \operatorname{Res}(g, e^{-i\theta}) = \frac{e^{in\theta} - e^{-in\theta}}{2i\sin\theta} = \frac{2i\sin(n\theta)}{2i\sin\theta} = \frac{\sin(n\theta)}{\sin\theta}. \]
3. Evaluation of the contour integral:
3.1. By the Cauchy Residue Theorem:\[ \frac{1}{2\pi i} \oint_{|z|=2} \frac{z^n}{z^2 - 2z\cos\theta + 1} \, dz = \sum_{j=1}^2 \operatorname{Res}(g, z_j) = \frac{\sin(n\theta)}{\sin\theta}. \]
4. Conclusion:\(\frac{1}{2\pi i} \int_{|z|=2} \frac{z^n}{1 - 2z\cos\theta + z^2} \, dz = \frac{\sin(n\theta)}{\sin\theta}\). Q.E.D.