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I am trying to show that

$$\frac{1}{2\pi i}\int_{\gamma(0,3)}\frac{e^{(zk)}}{z^2+1}\mathrm dz = \sin k$$ for all $k \in \mathbb{C}$. And $\gamma(z_0,R)$ is the circular contour $z_0+Re^{it}$, where $0\le t\le 2\pi.$ I know I need to use contour integration, but I cant seem to solve this problem.

Do I possibly have to use $z=e^{it}$ and $\sin t=\frac{1}{2i}(z-\frac{1}{z})?$ Its just a thought, I most probably might be wrong on this..

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1 Answer 1

Easier way to go would be to write $\frac{{{e^{zk}}}} {{{z^2} + 1}} = \frac{1} {{2i}}\left( {\frac{{{e^{zk}}}} {{z - i}} - \frac{{{e^{zk}}}} {{z + i}}} \right)$ and then noticing that both singularities are located inside of your contour of integration and that it is positively oriented. By applying residue theorem, you get $\frac{1} {{2\pi i}}\int_{\gamma \left( {0,3} \right)} {\frac{{{e^{zk}}}} {{{z^2} + 1}}dz} = \frac{{{e^{ik}} - {e^{ - ik}}}} {{2i}} = \frac{{\cos k + i\sin x - \cos \left( { - k} \right) - i\sin \left( { - k} \right)}} {{2i}} = \sin k$

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Thanks, shall work on it, I am not so confident with the residue theorem so I'll work on this! –  Thomas Jan 19 '12 at 0:52
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