Question: Evaluate the given integral using Cauchy’s Formula or Theorem: $\int_{|z|=2}\frac{e^z}{z(z-3)}dz$.
Answer:
For $z(z-3)=0$, $z=0$ or $z=3$, where only $z=0$ is bounded by $|z|=2$, thus $z_0=0$.
$\int_{|z|=2}\frac{e^z}{z(z-3)}dz = \int_{|z|=2}\frac{\frac{e^z}{z-3}}{z-0}dz$
(By Cauchy's formula)
$= 2\pi i f(z_0)$, where $f(z) = \frac{e^z}{z-3}$ is analytic on $\mathbb{C}$,
Thus, $2\pi i f(z_0) = 2\pi i \cdot \frac{e^0}{0-3} = -\frac{2\pi i}{3}$.
Therefore, $\int_{|z|=2}\frac{e^z}{z(z-3)}dz = -\frac{2\pi i}{3}$.
