Integral question

**Glitch** · June 16th 2011, 08:39 PM

The question
Evaluate $\oint_{|z| = 4} \frac{z^3 \ dz}{(z - 3)^3(z - 2)^2}$ using the Cauchy Integral Formula, where the circle is traversed once anti-clockwise.

My attempt
z = 3, z = 2 are both singularities in the contour, so:

Rewrite the integrand such that one of the singularities is in the denominator:
$\frac{\frac{z^3}{(z - 3)^3}}{(z - 2)^2}$

Thus by CIF, $f(z) = \frac{z^3}{(z - 3)^3}$
This singularity is of pole 2, so we must use $\frac{2\pi i}{1!} f^{1}(2)$ :
$2\pi i (\frac{3z^3(z - 3)^2) - 3z^2(z - 3)^3}{(z - 3)^6})$
= 72\pi i

Now for the other singularity:
$\frac{\frac{z^3}{(z - 2)^2}}{(z - 3)^3}$

Thus by CIF, $f(z) = \frac{z^3}{(z - 2)^2}$
This singularity is of pole 3, so we must use $\frac{2\pi i}{2!} f^{2}(3)$ :
$\pi i(\frac{6z^3-6z^2(z - 2) - 6z^2(z - 2) + 6z(z - 2)^2}{(z - 2)^4})$
= $72\pi i$

Add them together, and we get $144\pi i$ . However, the answer is actually 0. I think I'm supposed to be getting $-72\pi i$ for the second one, but I can't work out where I've gone wrong. :/

Any help would be greatly appreciated!

**TheEmptySet** · June 16th 2011, 08:55 PM

Hint use partial fractions first.

z^3/((z-3)^3(z-2)^2) - Wolfram|Alpha

**Glitch** · June 16th 2011, 09:31 PM

My lecturers notes say that using partial fractions is an alternative method to the one I've done. But looking now, he was using singularities of pole 1, so perhaps it's different?

Hmm, actually, just realised that my functions are not analytic in the contour, which breaks it. >_<

**chisigma** · June 16th 2011, 11:36 PM

The procedure has been described in...

http://www.mathhelpforum.com/math-he...la-183122.html

Here inside the contour f(z) has a second order pole in z=2 and a third order pole in z=3, so that what You have to do is compute the residues...

$r_{z=2} = \lim_{z \rightarrow 2} \frac{d}{d z} \{(z-2)^{2}\ f(z)\}$

$r_{z=3} = \lim_{z \rightarrow 3} \frac{1}{2}\ \frac{d^{2}}{d z^{2}} \{(z-3)^{3}\ f(z)\}$

... and the integral is...

$\int_{c} f(z)\ dz = 2 \pi i (r_{z=2} + r_{z=3})$

Kind regards

$\chi$ $\sigma$

**Aki** · June 17th 2011, 07:37 AM

Originally Posted by Glitch

This singularity is of pole 2, so we must use $\frac{2\pi i}{1!} f^{1}(2)$ :
$2\pi i (\frac{3z^3(z - 3)^2) - 3z^2(z - 3)^3}{(z - 3)^6})$
= 72\pi i

There is an error in the derivation.
The correct derivative is
$f'(z)=\frac{3z^2 (z-3)^3 - z^3 \cdot 3(z-3)^2}{(z-3)^6}=-\frac{9z^2}{(z-3)^4}.$
So, $f'(2)=-36.$

The derivation formula for $f(z)/g(z)$ is
$\frac{d}{dz}\frac{f(z)}{g(z)}=\frac{f'(z)g(z)-f(z)g'(z)}{g(z)^2}.$
In your equation, the numerator is replaced by $f(z)g'(z)-f'(z)g(z)$ .
It is interesting that you got the correct result for the second integral.
Does double use of the wrong derivation formula derive correct result ?

Sorry for my poor English.

**Glitch** · June 17th 2011, 07:41 AM

Huh, I did do it the wrong way around. That's quite worrying actually. :/

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