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Let C be the circle $|z| = 2$ traversed once counterclockwise. By using cauchy’s integral formula, compute $$\int_C \frac {sinz}{(z^2+1)^2}dz$$

I think that the singularities would be at i and -i, because $(z^2+1) = (z+i)(z-i)$. And both i and -i are in C, so I'm not sure how to deal with this. If the numerator was a polynomial, then I would do a partial fraction decomposition and solve it as a sum of integrals, but I can't do this with $sinz$.

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  • $\begingroup$ What is stopping you from doing a partial fraction decomposition? Can you do a partial fraction decomposition of $\frac{1}{(z^2+1)^2}$? If you multiply both sides by $\sin(z)$ what happens? $\endgroup$ – JMoravitz Mar 4 '16 at 3:58
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Hint:

$\frac{1}{(z^2+1)^2} = \frac{1}{(z-i)^2(z+i)^2} = \frac{i}{4(z+i)}-\frac{1}{4(z+i)^2}-\frac{i}{4(z-i)}-\frac{1}{4(z-i)^2}$

So:

$\frac{\sin(z)}{(z^2+1)^2} = \frac{i\sin(z)}{4(z+i)}-\frac{\sin(z)}{4(z+i)^2}-\frac{i\sin(z)}{4(z-i)}-\frac{\sin(z)}{4(z-i)^2}$

Continue as you originally planned.

For reference:

Cauchy's Integral Formula: for $f$ holomorphic over a closed disk with boundary $\gamma$, then for every $a$ in the interior of $\gamma$ one has $$f^{(n)}(a) = \frac{n!}{2\pi i}\int_\gamma \frac{f(z)}{(z-a)^{n+1}}dz$$

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