Let \(n\) be a positive integer and \(0 < \theta < \pi\) . Prove that
\[ { \frac { 1 } { 2 \pi i } } \int _ { | z | = 2 } { \frac { z ^ { n } } { 1 - 2 z \cos \theta + z ^ { 2 } } } d z = { \frac { \sin n \theta } { \sin \theta } } . \]
Solution
Set \[ F(z) = \frac{z^n}{1-2z\cos\theta+z^2}. \]
1. The denominator factors as\[ 1-2z\cos\theta+z^2 = (z-e^{i\theta})(z-e^{-i\theta}). \]
Since \[ e^{i\theta}+e^{-i\theta}=2\cos\theta \] and \[ e^{i\theta}e^{-i\theta}=1, \] expanding the right-hand side gives \[ z^2-2z\cos\theta+1. \]
2. The only poles of \(F\) are the distinct simple poles\[ e^{i\theta} \qquad\text{and}\qquad e^{-i\theta}, \]and both lie inside the contour \(\abs{z}=2\).
By step 1 these are the only zeros of the denominator. They are distinct because \(0<\theta<\pi\), and both have modulus \(1<2\).
3. The residue at \(z=e^{i\theta}\) is\[ \Res(F;e^{i\theta}) = \frac{e^{in\theta}}{2i\sin\theta}. \]
Using the factorization in step 1, \[ \begin{aligned} \Res(F;e^{i\theta}) &= \frac{e^{in\theta}} {e^{i\theta}-e^{-i\theta}}\\ &= \frac{e^{in\theta}}{2i\sin\theta}. \end{aligned} \]
4. The residue at \(z=e^{-i\theta}\) is\[ \Res(F;e^{-i\theta}) = -\frac{e^{-in\theta}}{2i\sin\theta}. \]
Again by step 1, \[ \begin{aligned} \Res(F;e^{-i\theta}) &= \frac{e^{-in\theta}} {e^{-i\theta}-e^{i\theta}}\\ &= -\frac{e^{-in\theta}}{2i\sin\theta}. \end{aligned} \]
5. The sum of the residues inside \(\abs{z}=2\) is\[ \frac{\sin(n\theta)}{\sin\theta}. \]
By steps 3 and 4, \[ \begin{aligned} \Res(F;e^{i\theta}) +\Res(F;e^{-i\theta}) &= \frac{e^{in\theta}-e^{-in\theta}} {2i\sin\theta}\\ &= \frac{\sin(n\theta)}{\sin\theta}. \end{aligned} \]
6. One has\[ \boxed{ \frac1{2\pi i} \int_{\abs{z}=2} \frac{z^n} {1-2z\cos\theta+z^2} \,dz = \frac{\sin(n\theta)}{\sin\theta}. } \]
4. The residue theorem and step 5 give\[ \int_{\abs{z}=2}F(z)\,dz = 2\pi i \frac{\sin(n\theta)}{\sin\theta}. \]Divide by \(2\pi i\).
7. Q.E.D.
Step 6 is the required identity.