Evaluate the given integral using Cauchy’s Formula or Theorem.
$$\int_{|z|=2} \frac{e^z \ dz}{z(z-3)}$$
First, we can find that $\frac{e^z \ dz}{z(z-3)}$ is not analytic when $z=0$ and $z=3$,
also, $z=3$ is outside the circle $|z|=2$ and $z=0$ is inside the circle $|z|=2$.
Hence, $$\int_{|z|=2} \frac{e^z \ dz}{z(z-3)} = \int_{|z|=2} \frac{ \frac{e^z }{z-3}}{z}dz$$
By Cauchy Formula, we can get $$f(z)= \frac{e^z}{z-3} , \ and \ z_{0} = 0$$
Therefore, $$\int_{|z|=2} \frac{e^z \ dz}{z(z-3)} = 2 \pi i f(z_0) =2 \pi i \frac{e^0}{0-3}\ = -\frac{2 \pi i}{3}$$
