Evaluate the integrals $$\int_{|z|=1}\dfrac{\cos z}{z-3}dz$$ and $$\int_{|z|=10}\dfrac{\cos z}{z-3}dz$$

The first one should be $0$, since the function $\dfrac{\cos z}{z-3}$ is holomorphic in the open disk $|z|<3$.

I've tried parametrizing $z=10e^{i\theta}$ for $\theta\in[0,2\pi]$. The second integral becomes $$\int_0^{2\pi}\dfrac{\cos (10e^{i\theta})}{10e^{i\theta}-3}\cdot 10ie^{i\theta}d\theta$$ and I don't know how to continue from here. Or perhaps I should use Cauchy's integral formula, which says that the integral is equal to $$2\pi i\cdot f(3)\cdot n(\gamma,3) = \cos(3)\cdot \int_{|z|=10}\dfrac{1}{z-3}dz$$ How can I integrate this last one?

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    $\begingroup$ Cauchy's integral formula says that your second integral equals $2\pi i\cos 3$. $\endgroup$ – njguliyev Oct 5 '13 at 18:58

Parametrizing the curve, etc., can sometimes work but it fails to exploit the efficiency afforded by Cauchy's discoveries in the $19$th century. Those advantages will enable you to compute integrals you'd never find via parametrizing and so on, reducing to real variables.

Cosine is an entire function, so this function is holomorphic except at the one point where the denominator is $0$. That point is not surrounded by your first curve, so the integral over that curve is $0$.

It is surrounded by your second curve, so the integral is equal to $\cos 3$ times the integral of $dz/(z-3)$ over any curve that winds once counterclockwise around $3$.


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