0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ To be clear: $C(0, 2)$ means a circle of radius 2 centered at the origin? $\endgroup$ – Sean Roberson May 20 '18 at 20:06
  • $\begingroup$ @SeanRoberson Yes! $\endgroup$ – weepyhollow May 20 '18 at 20:07
1
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ 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! $\endgroup$ – weepyhollow May 20 '18 at 20:39
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $e^{ix} = \cos x + i\sin x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.