Question: Evaluate the given integral using Cauchy's Formula or Theorem.
$$\int_{|z| = 2} \frac{e^z}{z(z-3)}dz$$
Answer:
We can find that on the region $|z| = 2$, $F(z) = \frac{e^z}{z(z-3)}$ not continuous at z = 0. Therefore I'll apply Cauchy's Formula.
$$\int_{|z| = 2} \frac{e^z}{z(z-3)}dz = \int_{|z| = 2} \frac{e^z /(z-3)}{z}dz$$
$$\implies f(z) = \frac{e^z}{z-3}, f(0) = -\frac{1}{3}$$
$$\int_{|z| = 2} \frac{e^z}{z(z-3)}dz = 2\pi i f(0) = -\frac{2\pi i}{3}$$
