# Problem 4.34

[4.34] Use the Cauchy Integral Formula(Theorem 4.30) to evaluate the integral in Exercise 4.33 when r=3.

Consider the integral $\int_{c[0,3]} \frac{1}{z^2-2z-8}dz=\int_{c[0,3]} \frac{dz}{(z+2)(z-4)}.$ Notice the function $$f(z)=\frac{1}{z-4}$$ is holomorphic in $$\mathbb{C}$$$$\setminus\left\{4\right\}$$ which contains $$D\overline [-2,3]$$. Thus, we can apply the Cauchy Integral Formula(Theorem 4.30): \begin{aligned} \frac{1}{2\pi i}\int_{c[-2,3]}\frac{\frac{1}{z-4}}{z+2}dz\\ =2\pi i f(-2)\\ =2\pi i (-\frac{1}{6})\\ = \frac{-\pi i}{3}\\\end{aligned} 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 $$C[0,3] \sim\mathbb{C}\setminus\left\{-2\right\} C[-2,3]$$ then $\int_{c[0,3]}\frac{dz}{z^2-2z-8}= \int_{c[-2,3]}\frac{dz}{z^2-2z-8}$ Therefore,$\int_{c[0,3]}\frac{dz}{z^2-2z-8}=\frac{-\pi i}{3}$