Using Cauchy Formula to find $\int_{|z|=2}\frac{e^z}{z(z-3)}dz$

Question

I am a bit confused on how I would use Cauchy's to find the integral:

$$\int_{|z|=2}\frac{e^z}{z(z-3)}dz$$

Any help would be appreciated!

Edit: $z = 0$ inside $z = 2$, so it is a singularity.

$$\int_{|z|=2}\frac {e^z}{z(z-3)}dz=\int_{|z|=2}\frac {e^z/(z-3)}{z}dz=2\pi i\cdot f(0)=2\pi i\cdot -\frac{1}{3}=-\frac{2\pi i}{3}.$$

Does this look good? :o

Answer 1

Yes, now your work is correct the answer is $-\frac{2\pi i}{3}$.

The mapping $f:z\mapsto \frac{e^{z}}{z-3}$ is holomorphic everywhere in the interior of $|z|=2$ so by Cauchy Integral formula, we have $$\int_{|z|=2}\frac{e^{z}}{z(z-3)}\, {\rm d}z=\int_{|z|=2}\frac{\frac{e^{z}}{z-3}}{z}\, {\rm d}z=2\pi if(0)=2\pi i\left(\frac{e^{0}}{0-3}\right)=-\frac{2\pi i}{3}.$$

Recall that if $f: \Omega\subseteq \mathbb{C}\to \mathbb{C}$ is holomorphic function and $\gamma$ is a circle contained in $\Omega$. Then for all $z_{0}$ in the disk bounded by $\gamma$, $$f^{(n)}(z_{0})=\frac{n!}{2\pi i}\int_{\gamma}\frac{f(z)}{(z-z_{0})^{n+1}}\, {\rm d}z.$$