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I have the problem, to use Cauchy's theorem or Cauchy's integral formula to evaluate $\oint_C \dfrac{\sin2z}{6z-\pi} dz$, where $C$ is the circle $\mod (z) = 3$.

I know that the denominator vanishes at z = $\dfrac{\pi}{6}$, which is within the contour, so the result of the integral is not $0$ (Cauchy's theorem). I have a singularity within the contour, then, and it is not analytic. How then can I use Cauchy's theorem or integral formula for this?

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2 Answers 2

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No need for the residue theorem, Cauchy's integral formula gives the answer right away:

$$ \oint_C \dfrac{\sin2z}{6z-\pi}\,dz = \oint_C \dfrac{\frac{\sin2z}{6}}{z-\frac{\pi}{6}}\,dz = 2\pi i\,\frac{\sin \frac{\pi}3}{6} = \frac{\pi i\sqrt{3}}{6}. $$

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how does Cauchy's integral formula give this? –  user1535776 Feb 4 '13 at 22:00
    
@user1535776 Which step is unclear? –  mrf Feb 4 '13 at 22:07
    
Why does $(z - \dfrac{\pi}{6})$ go away, and how does $2 \pi i$ come about? It looks like residue theorem, how did you get it from Cauchy's integral formula? –  user1535776 Feb 4 '13 at 22:25
    
@user1535776 Cauchy's integral formula says that $$\oint_C \frac{f(z)}{z-a}\,dz = 2\pi i f(a)$$ if $f$ is analytic on and inside $C$. Here $a=\pi/6$ and $f(z)=\frac16 \sin 2z$. –  mrf Feb 4 '13 at 22:33
    
Okay, perfect, thanks! –  user1535776 Feb 4 '13 at 22:34

Apply the Residue Theorem. The integral has a singularity at $z=\frac{\pi}{6}$, which is inside the disk, so the value is $$2\pi i \text{Res} \left(\frac{1}{6}\frac{\sin (2z)}{z-\pi/6},\ z=\frac{\pi}{6}\right).$$

This will equal $$\frac{\pi i}{3}\sin\left(\frac{\pi}{3}\right)= \frac{\pi i}{2\sqrt{3}}.$$

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Don't know why I missed that, perfect –  user1535776 Feb 4 '13 at 0:17

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