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$$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z$$

Well i know by a fundamental result that:

$$\int_{\gamma(a;r)}\frac{1}{z-a} \mathrm{d}z=2\pi\imath$$

But here the point $$z=-2$$ lies outside $$\gamma(0;1)$$ so what is the reasoning behind could you help me please?