Evaluate the integral using Cauchy Integral Formula
$\displaystyle \oint\frac{e^z}{z(z^2-9)}$
with $\displaystyle |z| = 1$
Note $\displaystyle z=\pm 3$ is outside of $\displaystyle \left|z\right|=1$. The only pole is the simple pole $\displaystyle z=0$ because its contained in $\displaystyle \left|z\right|=1$.
So now we find the Residue:
$\displaystyle \text{Res}\left(0\right)=\lim_{z\to 0}z\left(\frac{e^z}{z(z^2-9)}\right)=-\frac{1}{9}$.
Thus, by the Cauchy Residue Theorem, $\displaystyle \oint_{\left|z\right|=1}\frac{e^z}{z(z^2-9)}\,dz=2\pi i \text{Res}\left(0\right)=-\frac{2\pi i}{9}$
Does this make sense?
Sorry, my mistake. I was using Cauchy's Residue Theorem instead of Cauchy's Integral Theorem.
You'll get the same result in the end.
Note that $\displaystyle \oint_{\left|z\right|=1}\frac{e^z}{z(z^2-9)}\,dz=\oint_{\left|z\right|=1}\frac{e^z/\left(z^2-9\right)}{z-0}\,dz$. Let $\displaystyle f\!\left(z\right)=\frac{e^z}{z^2-9}$ (since $\displaystyle z^2-9$ has no zeros in |z|=1)and let $\displaystyle z_0=0$. Then by Cauchy's Integral Theorem, we have $\displaystyle \oint_{\Gamma}\frac{f\!\left(z\right)}{z-z_0}\,dz=2\pi i f\!\left(z_0\right)\implies \oint_{\left|z\right|=1}\frac{e^z/\left(z^2-9\right)}{z-0}\,dz=2\pi i \left(\frac{e^0}{0^2-9}\right)=-\frac{2\pi i}{9}$.
I hope this helps.