1. ## Cauchy Integral

Evaluate the integral using Cauchy Integral Formula

$\oint\frac{e^z}{z(z^2-9)}$

with $|z| = 1$

2. Originally Posted by Richmond
Evaluate the integral using Cauchy Integral Formula

$\oint\frac{e^z}{z(z^2-9)}$

with $|z| = 1$
Note $z=\pm 3$ is outside of $\left|z\right|=1$. The only pole is the simple pole $z=0$ because its contained in $\left|z\right|=1$.

So now we find the Residue:

$\text{Res}\left(0\right)=\lim_{z\to 0}z\left(\frac{e^z}{z(z^2-9)}\right)=-\frac{1}{9}$.

Thus, by the Cauchy Residue Theorem, $\oint_{\left|z\right|=1}\frac{e^z}{z(z^2-9)}\,dz=2\pi i \text{Res}\left(0\right)=-\frac{2\pi i}{9}$

Does this make sense?

3. Isnt this the same as Residue Theorem?

So in other words, when it is stated to use Cauchy Integral Formula, we find the poles and see which pole are contained within z and use the residue theorem to evaluate the integral?

Anyway, thanks alot.

4. Originally Posted by Richmond
Isnt this the same as Residue Theorem?

So in other words, when it is stated to use Cauchy Integral Formula, we find the poles and see which pole are contained within z and use the residue theorem to evaluate the integral?

Anyway, thanks alot.
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 $\oint_{\left|z\right|=1}\frac{e^z}{z(z^2-9)}\,dz=\oint_{\left|z\right|=1}\frac{e^z/\left(z^2-9\right)}{z-0}\,dz$. Let $f\!\left(z\right)=\frac{e^z}{z^2-9}$ (since $z^2-9$ has no zeros in |z|=1)and let $z_0=0$. Then by Cauchy's Integral Theorem, we have $\oint_{\Gamma}\frac{f\!\left(z\right)}{z-z_0}\,dz=2\pi i f\!\left(z_0\right)\implies \oint_{\left|z\right|=1}\frac{e^z/\left(z^2-9\right)}{z-0}\,dz=2\pi i \left(\frac{e^0}{0^2-9}\right)=-\frac{2\pi i}{9}$.

I hope this helps.

5. yea, thanks alot for the help. =)