# Another Cauchy integral

• Jul 12th 2009, 09:23 AM
Richmond
Another Cauchy integral
Another question to be done using Cauchy's integral formula:

$\displaystyle \oint\frac{e^z}{z(z^2-9)} \, {\color{red}dz}$ with $\displaystyle |z-3|=1$

Start by letting $\displaystyle f(z) = \frac {e^z}{z}$ since z has no zeros in $\displaystyle |z-3|=1$

then $\displaystyle \oint\frac{f(z)}{(z^2-9) - z_0}$ with $\displaystyle z_0 = +-3$

So $\displaystyle 2\pi i * f(z_0) = 2\pi i*\frac{e^3}{3}$ and $\displaystyle 2\pi i*\frac{e^{-3}}{-3}$

Did i go wrong anywhere?
• Jul 12th 2009, 06:53 PM
mr fantastic
Quote:

Originally Posted by Richmond
Another similiar question,

$\displaystyle \oint\frac{e^z}{z(z^2-9)} \, {\color{red}dz}$ with $\displaystyle |z-3|=1$

Start by letting $\displaystyle f(z) = \frac {e^z}{z}$ since z has no zeros in $\displaystyle |z-3|=1$

then $\displaystyle \oint\frac{f(z)}{(z^2-9) - z_0}$ with $\displaystyle z_0 = +-3$

So $\displaystyle 2\pi i * f(z_0) = 2\pi i*\frac{e^3}{3}$ and $\displaystyle 2\pi i*\frac{e^{-3}}{-3}$

Did i go wrong anywhere?

The contour encloses only one singularity: A simple pole at z = 3. So take $\displaystyle f(z) = \frac{e^z}{z(z + 3)}$.