cauchy integral formula

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June 2nd 2010, 08:10 PM
scorpion007

cauchy integral formula

Eval

$\frac{1}{2\pi i}\oint\frac{e^{tz}}{(z^2+1)^2}dz$ if t>0 and C is the circle {|z|=3}

Hmm...

Am I suppose to parameterise something?

I can write for example z(t)=3e^{it} then dz = 3i e^{it}dt and sub it back in, but that just makes it messy and complicated.

Am I missing an easier way?
June 2nd 2010, 08:23 PM
Drexel28

Quote:

Originally Posted by scorpion007

Eval

$\frac{1}{2\pi i}\oint\frac{e^{tz}}{(z^2+1)^2}dz$ if t>0 and C is the circle {|z|=3}

Hmm...

Am I suppose to parameterise something?

I can write for example z(t)=3e^{it} then dz = 3i e^{it}dt and sub it back in, but that just makes it messy and complicated.

Am I missing an easier way?

Do you know the Residue theorem? You're function has singularities at $z=\pm i$ both of which lie in $C$ . So, $\oint_{C}f(z)dz=\underset{z= i}{\text{Res}}f(z)+\underset{z=- i}{\text{Res}}f(z)$
June 2nd 2010, 08:26 PM
scorpion007

Thanks, I considered that just now, but this question belongs to a section well-prior to covering the residue thm, in the CIF section. So I suppose it should be solvable using that...
June 2nd 2010, 08:36 PM
Drexel28

Quote:

Originally Posted by scorpion007

Thanks, I considered that just now, but this question belongs to a section well-prior to covering the residue thm, in the CIF section. So I suppose it should be solvable using that...

Ok, decompose your the denominator into $(z-i)^2(z+i)^2$ and decompose this using partial fractions and then from there apply Cauchy's integral theorem.
June 2nd 2010, 09:35 PM
scorpion007

I get

$-\frac{t}{2} \cos t$

but the answer says: $1/2 (\sin t -t \cos t)$

for the partial fractions I got A = B = -1/4 e^{tz}

so we have

$\frac{1}{2\pi i} \oint_C\frac{e^{tz}}{(z^2+1)^2}dz= $

$\frac{1}{2\pi i} \left\{-\left(\oint_C\frac{e^{tz}/4}{(z+i)^2}\; dz + \oint_C\frac{e^{tz}/4}{(z-i)^2}\;dz\right)\right\}=$

$-\frac{1}{2\pi i}2\pi i \frac{1}{4}(f'(-i)+f'(i))$ where

$f(z)=e^{tz}$ and $f'(z)=te^{tz}$

$=-\frac{t}{4}(e^{-it}+e^{it})$

$=-\frac{t}{4}(2\cos t)$

$=-\frac{t}{2}\cos t$

Did I make a mistake?
June 2nd 2010, 09:55 PM
Drexel28

Quote:

Originally Posted by scorpion007

I get

$-\frac{t}{2} \cos t$

but the answer says: $1/2 (\sin t -t \cos t)$

for the partial fractions I got A = B = -1/4 e^{tz}

so we have

$\frac{1}{2\pi i} \oint_C\frac{e^{tz}}{(z^2+1)^2}dz= $

$\frac{1}{2\pi i} \left\{-\left(\oint_C\frac{e^{tz}/4}{(z+i)^2}\; dz + \oint_C\frac{e^{tz}/4}{(z-i)^2}\;dz\right)\right\}=$

$-\frac{1}{2\pi i}2\pi i \frac{1}{4}(f'(-i)+f'(i))$ where

$f(z)=e^{tz}$ and $f'(z)=te^{tz}$

$=-\frac{t}{4}(e^{-it}+e^{it})$

$=-\frac{t}{4}(2\cos t)$

$=-\frac{t}{2}\cos t$

Did I make a mistake?

I don't think you decomposed it correct.
June 2nd 2010, 10:00 PM
scorpion007

ah yes! thanks!