# Integral question

• Jun 16th 2011, 08:39 PM
Glitch
Integral question
The question
Evaluate $\displaystyle \oint_{|z| = 4} \frac{z^3 \ dz}{(z - 3)^3(z - 2)^2}$ 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:
$\displaystyle \frac{\frac{z^3}{(z - 3)^3}}{(z - 2)^2}$

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

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

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

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

Any help would be greatly appreciated!
• Jun 16th 2011, 08:55 PM
TheEmptySet
Re: Integral question
• Jun 16th 2011, 09:31 PM
Glitch
Re: Integral question
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. >_<
• Jun 16th 2011, 11:36 PM
chisigma
Re: Integral question
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...

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

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

... and the integral is...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jun 17th 2011, 07:37 AM
Aki
Re: Integral question
Quote:

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

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

The derivation formula for $\displaystyle f(z)/g(z)$is
$\displaystyle \frac{d}{dz}\frac{f(z)}{g(z)}=\frac{f'(z)g(z)-f(z)g'(z)}{g(z)^2}.$
In your equation, the numerator is replaced by $\displaystyle f(z)g'(z)-f'(z)g(z)$.
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.
• Jun 17th 2011, 07:41 AM
Glitch
Re: Integral question
Huh, I did do it the wrong way around. That's quite worrying actually. :/