Contour Integration

**adam_leeds** · May 12th 2011, 11:24 AM

$\int_c\frac{{e}^{z } }{ ({{z}^{2 }-9 })^{2 } } dz$ where c is the positively oriented circle $|z+2|=2$ So the formula is $\frac{2\pi i {f}^{(n-1) } (z)}{(n-1)! }$ $f(z) = {e}^{ z} and n = 2$ So my answer is $2\pi i {e}^{-3 }$ But according to my teacher it is over 27

**Amer** · May 12th 2011, 11:29 AM

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}

**adam_leeds** · May 12th 2011, 11:35 AM

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

**adam_leeds** · May 12th 2011, 11:39 AM

Originally Posted by adam_leeds

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

Sorry still dont get it.

**Mondreus** · May 12th 2011, 12:31 PM

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

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

$f'(-3)=\frac{1!}{2\pi i}\int_{|z+2|=2}\frac{f(z)}{(z+3)^2}dz=\frac{1!}{2 \pi i}\int_{|z+2|=2}g(z)dz$

**Amer** · May 12th 2011, 12:32 PM

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

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

$\frac{\frac{e^z}{(z-3)^2}}{(z+3)^2}$ the function is

$\frac{f(z)}{(z+3)^2}$

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