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Let $\gamma(z_0,R)$ denote the circular contour $z_0+Re^{it}$ for $0\leq t\leq 2\pi$. Evaluate: $\int_{\gamma(0,1)} \frac{\cos(z)}{z^2}dz$

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    What have you tried? Cauchy's integral formula or the residue theorem work well here.2012-12-20

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I don't know what you mean by $\gamma(0,1)$ but I will assume you want the closed line integral over $\gamma$. If $D=\left\{\left|z-z_0\right| then:

If $0\in D$, by Cauchy's differentiation formula, $\oint_{\gamma}\frac{\cos z}{z^2}dz=2\pi i f^{\prime}(0)$ where $f(z)=\cos z$

If $0\notin D$ then $0\notin \gamma([0,2\pi])$ (so that your integral is defined) and by Cauchy's Integral Theorem, $\oint_{\gamma}\frac{\cos z}{z^2}dz=0$ ($f$ is analytic in $D$)

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    @Fabian Yeah I will edit my answer2012-12-20
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${\cos(z)\over z^2}={1\over z^2}(1-{z^2\over 2!}+{z^4\over 4!}-...)={1\over z^2}-{1\over 2!}+{z^2\over 4!}-...\implies$ Coefficient of ${1\over z}$ in the Laurent series expansion of ${\cos(z)\over z^2}=0\implies$ $\int_{\gamma(0,1)} \frac{\cos(z)}{z^2}dz=2\pi i\times0=0.$