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\end{align*}  Hence,   $$\int_{c[-2,3]} \frac{dz}{z^2-2z-8}=\frac{-\pi i}{3}$$  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   $$\int_{c[0,3]} \frac{dz}{z^2-2z-8=\int_{c[-2,3]}\frac{dz}{z^2-2z-8}$$