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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. – Ragib Zaman Dec 20 '12 at 14:39

${\cos(z)\over z^2}={1\over z^2}(1-{z^2\over 2!}+{z^4\over 4!}-...)={1\over z^2}-{1\over 2!}+{z^2\over 4!}-...\implies$ Coefficient of ${1\over z}$ in the Laurent series expansion of ${\cos(z)\over z^2}=0\implies$ $\int_{\gamma(0,1)} \frac{\cos(z)}{z^2}dz=2\pi i\times0=0.$

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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 $D=\left\{\left|z-z_0\right|<R\right\}$ 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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Shouldn't it be $f(z) = \cos z$? (and thus your distinction into cases isn't really necessary) – Fabian Dec 20 '12 at 15:48
@Fabian Yeah I will edit my answer – Nameless Dec 20 '12 at 16:47

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