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I'm trying to solve the complex integral $\int \sin(z)dz$ over the contour shown below.

enter image description here

The approach I took was to parametrize the semicircle and the line separately, obtaining $z(t)= 3e^{it}$ and $z(t)=3+3t+3it$ respectively.

However, when I'm going to solve the integrals these two functions become the argument of the $\sin(z)$ function. I don't know how to proceed form here. Am I doing anything wrong? Is there a better way to solve it?

Eric Angle
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Enrique
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    $\sin(z)$ is analytic with antiderivative $-\cos(z)$. The integral depends only on the endpoints of the contour and hence $\int_{z_1}^{z_2} \sin(z),dz=\cos(z_1)-\cos(z_2)$ – Mark Viola Mar 19 '17 at 19:13
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    Dr. MV´s "comment" is, in fact, a complete and accurate answer. Analicity is a very powerful characteristic some complex functions have, and when they do sometimes things can get extremely easy. – DonAntonio Mar 19 '17 at 19:38
  • Oh I understand! Thank you both for your help! – Enrique Mar 20 '17 at 00:38

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