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Show that

$$\frac{1}{2\pi i}\int_\gamma \frac{z^n}{1-2z \cos\theta +z^2} dz = \frac{\sin(n\theta)}{\sin\theta}$$, where

$$n \in Z_{\ge1}\;.\;\;\theta \in (0,\pi)\,,\;\ \gamma = C(0,2)$$

traversed counterclockwise.

I'm not really sure how to approach this. I have tried substituting $cos\theta$ for $\frac{1}{2} (z+\frac{1}{z})$ to get $\frac{z^n}{2}$ inside the integral, but isn't that analytic, so the integral of that would be 0?

The solution states that the roots of the denominator are $e^{i\theta}$ and $e^{-i\theta}$ of order 1 and then to proceed using Cauchy's Integral Formula. I don't understand how this was reached at all. Any help would be appreciated!

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  • $\begingroup$ To be clear: $C(0, 2)$ means a circle of radius 2 centered at the origin? $\endgroup$ May 20, 2018 at 20:06
  • $\begingroup$ @SeanRoberson Yes! $\endgroup$ May 20, 2018 at 20:07

3 Answers 3

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Observe that the quadratic's discriminant is

$$\Delta=4\cos^2\theta-4=-4\sin^2\theta\implies z_{1,2}=\frac{2\cos\theta\pm\sqrt\Delta}2=\cos\theta\pm i\sin\theta=e^{\pm i\theta}$$

Thus:

$$I\;\;Res_{z=e^{i\theta}}(f)=\lim_{x\to e^{i\theta}}\frac{(z-e^{i\theta})z^n}{(z-e^{i\theta})(z-e^{-i\theta})}=\frac{e^{ni\theta}}{e^{i\theta}-e^{-i\theta}}=\frac{e^{ni\theta}}{2i\sin\theta}$$

$$II\;\;Res_{z=e^{-i\theta}}(f)=\lim_{x\to e^{-i\theta}}\frac{(z-e^{-i\theta})z^n}{(z-e^{i\theta})(z-e^{-i\theta})}=\frac{e^{-ni\theta}}{e^{-i\theta}-e^{i\theta}}=\frac{e^{-ni\theta}}{-2i\sin\theta}$$

By the Residue Theorem (can you see where, how and why?), the integral equals

$$I+II\;=\;\frac{e^{ni\theta}-e^{-ni\theta}}{2i\sin\theta}=\frac{\sin n\theta}{\sin\theta}$$

Note: we used above that

$$t\in\Bbb R\implies \sin t=\frac{e^{it}-e^{-it}}{2i}$$

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    $\begingroup$ This is perfect - I was able to proceed with Μάρκος Καραμέρης's hint but then I couldn't get the final answer because I was missing a negative sign, and this answer helped me resolve that issue! $\endgroup$ May 20, 2018 at 20:39
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Since the denominator has real coefficients you can see that the roots are $cos\theta+isin\theta$ and $cos\theta-isin\theta$ which are $e^{i\theta}$ and $e^{-i\theta}$. Cauchys formula now provides the answer.

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I think the easiest way to see this is to rewrite the integrand with a partial fraction decomposition. Since $1-2z\cos\theta + z^2 = (z-e^{i\theta})(z-e^{-i\theta})$, we have $$ \frac{1}{1-2z\cos\theta + z^2} = \frac{1/(2i\sin\theta)}{z-e^{i\theta}}+\frac{-1/(2i\sin\theta)}{z-e^{i\theta}}, $$ and then rewrite the original integral: $$ \frac{1}{2i\sin\theta}\cdot\frac{1}{2\pi i}\oint_\gamma\frac{z^n}{z-e^{i\theta}}\,dz - \frac{1}{2i\sin\theta}\cdot\frac{1}{2\pi i}\oint_\gamma\frac{z^n}{z-e^{-i\theta}}\,dz. $$ Applying Cauchy to the holomorphic function $f(z) = z^n$ in both integrals, we get $$ \frac{e^{i\theta n}}{2i\sin\theta} - \frac{e^{-i\theta n}}{2i\sin\theta} = \frac{1}{\sin\theta}\bigg[\frac{e^{i\theta n}-e^{-i\theta n}}{2i}\bigg] = \frac{\sin (n\theta)}{\sin \theta}, $$ where the last equality is using $e^{ix} = \cos x + i\sin x$.

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