4.34

4.34
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Use the Cauchy integral formula to evaluate the integral \(\int_{C[0,3]}\frac{dz}{z^2-2z-8}\) \[\begin{aligned} \int_{C[0,3]}\frac{dz}{z^2-2z-8}&=\int_{C[0,3]}\frac{dz}{(z+2)(z-4)}\\ &=\int_{C[0,3]}\frac{\frac{1}{z-4}}{z+2}\end{aligned}\] Assume \(C[0,3]\) and \(C[-2,2]\) are counter-clockwise oriented circles, then \(-2\) is inside both of the circles. By problem 4.21, we know that \(C[0,3]\backsim_{\mathbb{C} / \lbrace -2 \rbrace} C[-2,3]\). Then by Cauchy’s Theorem, in region \(\mathbb{C}\ \lbrace -2 \rbrace\) we have: \[\int_{C[0,3]}\frac{\frac{1}{z-4}}{z+2}=\int_{C[-2,3]}\frac{\frac{1}{z-4}}{z+2}\] Now let \(f(z)=\frac{1}{z-4}\). Then \(f\) is holomorphic in \(\overline{D}[02,3]\). So by Cauchy’s integral formula: \[\begin{aligned} \int_{C[0,3]}\frac{dz}{z^2-2z-8}&=\int_{C[-2,3]}\frac{\frac{1}{z-4}}{z+2}\\ &=(2\pi i)f(-2)\\ &=(2 \pi i)(-\frac{1}{6}\\ &=\frac{-\pi i}{3}\end{aligned}\]