# Contour Integration

• May 12th 2011, 11:24 AM
Contour Integration
$\displaystyle \int_c\frac{{e}^{z } }{ ({{z}^{2 }-9 })^{2 } } dz$where c is the positively oriented circle $\displaystyle |z+2|=2$ So the formula is $\displaystyle \frac{2\pi i {f}^{(n-1) } (z)}{(n-1)! }$ $\displaystyle f(z) = {e}^{ z} and n = 2$ So my answer is $\displaystyle 2\pi i {e}^{-3 }$ But according to my teacher it is over 27 (Headbang)
• May 12th 2011, 11:29 AM
Amer
f(z) is not e^z in this question

\frac{e^z}{(z^2-3)^(z^2+3)^2 } = \frac{\frac{e^z}{(z+3)^2}}{(z-3)^2}
• May 12th 2011, 11:35 AM
$\displaystyle \frac{e^z}{(z^2-3)^(z^2+3)^2 } = \frac{\frac{e^z}{(z+3)^2}}{(z-3)^2}$
• May 12th 2011, 11:39 AM
Quote:

$\displaystyle \frac{e^z}{(z^2-3)^(z^2+3)^2 } = \frac{\frac{e^z}{(z+3)^2}}{(z-3)^2}$

Sorry still dont get it.
• May 12th 2011, 12:31 PM
Mondreus
$\displaystyle g(z)=\frac{e^z}{(z-3)^2(z+3)^2}$

As you can see, only one of the poles (z=-3) lies within the given circle, so we let $\displaystyle f(z)=\frac{e^z}{(z-3)^2}$ and use Cauchy's integral formula for f(-3):

$\displaystyle f'(-3)=\frac{1!}{2\pi i}\int_{|z+2|=2}\frac{f(z)}{(z+3)^2}dz=\frac{1!}{2 \pi i}\int_{|z+2|=2}g(z)dz$
• May 12th 2011, 12:32 PM
Amer
http://latex.codecogs.com/png.latex?...z-3)^2(z+3)^2}

we have two singular points -3,3 but 3 is outside the circle
but -3 is inside so the singular point for the function is -3

http://latex.codecogs.com/png.latex?...)^2}}{(z+3)^2} the function is

http://latex.codecogs.com/png.latex?...f(z)}{(z+3)^2}