beginner cauchy integral equations

**punkstart** · August 9th 2011, 04:29 AM

all circles positively oriented. Use Cauchy's Integration Formulas to compute each of the following integrals. HINT: use partial fractions where appropriate.

a) $\int_{|z-2|=2} \frac{cosz}{(z-3)^2(z+1)}dz$
b) $\int_{|z+2|=1} \frac{sinz}{z^2(z-1)^2}dz$
c) $\int_{|z+2|=2} \frac{ze^{zt}}{(z+1)^3}dz$ , t>0

The Cauchy integral formula : $\int_C \frac{f(z)dz}{z-z_o}=2\pi i f(z_o)$

NOW FOR c) in the denominator -1 is interior to C so i go as follows,

$\int_{|z+2|=2} \frac{ze^{zt}}{(z+1)^3}dz=\int_{|z+2|=2} \frac{ze^{zt}}{(z-(-1))^{2+1}}dz$ so n=2 in the extension of Cauchy's integral formula given as $\int_C \frac{f(z)dz}{(z-z_0)^{n+1}}=\frac{2\pi i}{n!}f^{(n)}(z_0)$

so i let f(z) = $ze^{zt}$ then f'(z) = $e^{zt}+zte^{zt}$ and f ''(z) = $2te^{zt}+zt^2e^{zt} = e^{zt}(2t+zt^2)$
hence f '' (-1) = $e^{-t}(2t-t^2)$ and then by the extended Cauchy integral formula i get
$\int_{|z+2|=2} \frac{ze^{zt}}{(z+1)^3}dz = \frac{2\pi i}{2} e^{-t}(2t-t^2)= \pi i e^{-t}(2t-t^2)$ , is this correct ?

OK, things i don't know :

Now in a) neither 3 nor -1 are interior to C so neither of them are my Zo point. The same goes for b) in the denominator neither 0 nor 1 are interior to C, so where do i find Zo ? Now if the answer is to use partial fraction decomposition, my attempt :
for a) $\frac{cosz}{(z-3)^2(z+1)}= \frac{A}{z-3}+\frac{B}{(z-3)^2}+\frac{C}{z+1}$
$cosz = A(z-3)(z+1)+B(z+1)+C(z-3)^2$
$cosz=z^2(A+C)+z(-2A+B-6C)+(-3A+B+9C)$ and now what ? How do i equate anything??? I have the same confusion for b). I aprreciate any help.

**FernandoRevilla** · August 9th 2011, 05:10 AM

Originally Posted by punkstart

Now in a) neither 3 nor -1 are interior to C

That is not true, $3$ is (geometrically) interior to $C$ and $-1$ is (geometrically and topollogically) exterior to $C$ . So, choosing $f(z)=\cos z/(z+1)$ (analytic in $C$ and its geometrical interior) we have:

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

**punkstart** · August 9th 2011, 06:08 AM

I'm such a fool!! I apologise. But for b) i do believe i have that same problem, finding Zo? And how do i get partial fractions ?

**FernandoRevilla** · August 9th 2011, 06:26 AM

Originally Posted by punkstart

I'm such a fool!! I apologise. But for b) i do believe i have that same problem, finding Zo? And how do i get partial fractions ?

$f(z)=\dfrac{\sin z}{z^2(z-1)^2}$ is analytic on $C\equiv |z+2|=1$ and in its geometrical interior. So, by Cauchy's Theorem $\int_{C}f(z)=0$ .

**punkstart** · August 9th 2011, 07:14 AM

I consider myself a fool TWICE ! Much appreciated.

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