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Show that $\int_{C_1}f=\int_{C_2}f$, where $C_1:|z|=1$, $C_2:|z|=2$, and $f(z)=\frac{2z+1}{\sin z}$.

Hint: Locate the singularities of $f$ in each case and indicate their location with respect to the two given contours.

I think I know how to locate the singularities by setting the denominator equal to zero but I am not sure how to indicate their locations.

Community
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Katie Shaw
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  • After you locate the singularities like you said, state which ones lie in $C_1$ and which ones lie in $C_2$. Cauchy's residue theorem implies that if $C_1$ and $C_2$ contain (within their interior) the same isolated singularities, then the contour integrals are the same. – Taylor Jun 14 '15 at 15:33

2 Answers2

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We have sin z=z.infinite product_(n not 0) (1-z/(pi.n))e^z/(pi.n). So the only singularities of the denominator are n.pi, n is an integer. We see there are no singularities between the two circles and so the integrals are the same.

Adelafif
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HINT:

$$\sin z=0\implies z=0,\pm \pi, \pm 2\pi, \pm 3\pi, \cdots$$

and $-\pi<-2<-1<0<1<2<\pi$.

Mark Viola
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