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QUESTION

Find $\int_{bunny} {\cos z dz\over z^2(z-1)}$,\ \  (where bunny
means the boundary of the bunny below taken counterclockwise.)
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ANSWER

The function $\cos z dz\over z^2(z-1)$ has singularities at $z=0$
and $z=1$. Only $0$ lies inside bunny. So let $f(z)={\cos z\over
z-1}$ and then $f(z)\over z^2$= $\cos z dz\over z^2(z-1)$, where
$f$ is analytic within bunny and on the boundary of bunny. Thus by
the Cauchy integral formula $$\int_{bunny}  {\cos z dz\over
z^2(z-1)}=\int_{bunny}{f(z)\over z^2}=2\pi i f^{'}(0)=-2\pi i$$


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