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Veva Garcia edited untitled.tex
about 8 years ago
Commit id: 08c195388c26f34e993d077a41202d91b0fd1360
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By Theorem 4.23 (Cauchy's Theorem): if f is holomorphic in $\mathbb{C}\setminus\left\{-2\right\}$ and $C[0,3] ~\mathbb{C}\setminus\left\{-2\right\} C[-2,3]$
then
$$\int_{c[0,3]}\frac{dz}{z^2-2z-8}= \int_{c[-2,3]}\frac{dz}{z^2-2z-8} $$
Therefore,$$\int_{c[0,3]}\frac{dz}{z^2-2z-8=\frac{\pi Therefore,$$\int_{c[0,3]}\frac{dz}{z^2-2z-8}=\frac{-\pi i}{3}$$