Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I wish to calculate $\oint_{C}\frac{dz}{z(z-1)(z-2)}$ when $C$ is a circle around the origin with radius $1.5$.

I guess that I should somehow apply Cauchy's integral formula here, but $\frac{1}{z},\frac{1}{z-1}$ are not analytical inside of $C$ so I can't define something like $f(z)=\frac{1}{z(z-2)}$ and calculate $\oint_{C}\frac{f(z)dz}{(z-1)}$ by Cauchy's.

Can someone please help me understand how to calculate this integral ?

I am guessing there is some trick so I can use Cauchy's integral formula, but I didn't manage to think of any such tricks.

share|improve this question
    
Split your circle into two curves, each one surrounding just one singularity (or use the residue theorem). –  mrf Jan 10 '13 at 23:32

1 Answer 1

up vote 3 down vote accepted

$$ \oint_c \frac{1}{z(z-1)(z-2)} dz = 2\pi i \; \left [ \; \text{Res}\left (f(z), 0\right ) + \text{Res}\left ( f(z), 1 \right ) \; \right ] = 2\pi i(1/2 - 1) = -\pi i $$

Or simply, since pole of $z=2 > 1.5 $

$$ \oint_c \frac{1}{z(z-1)(z-2)} dz = \oint_c \left( \frac{1}{2 (-2+z)}-\frac{1}{-1+z}+\frac{1}{2 z} \right ) dz \\ = \oint_c \frac{1}{2z}dz - \oint_c\frac{1}{z-1} = \pi i - 2 \pi i = -\pi i$$

I hope I am not wrong!!

share|improve this answer
    
Thanks! How did you calculate the last two integrals (last line) ? –  Belgi Jan 11 '13 at 15:01
    
@Belgi check out this and this –  Santosh Linkha Jan 11 '13 at 22:29
    
Thanks, you are very kind! By the way: did you use the standard way of finding how to write it with partial fractions ? (or maybe you had a shortcut here ?) –  Belgi Jan 12 '13 at 0:55
    
@Belgi lol ... no, the standard way is to use Residue theorem, since i don't think we will be able to make use of partial fraction every time. I wrote it here since it was possible ... and it's easy to understand it. –  Santosh Linkha Jan 12 '13 at 0:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.