How to Evaluate a Cauchy Integral on Different Paths?

[redshift]

Cauchy integral question

The question calls for finding the integral of dz/((z-i)(z+1)) (C:|z-i|=1)

I can't figure out how to do this for (C:|z-i|=1). How does this differ from, say, (C: |z|=2)

Regards

 

[Galileo]

One way to evaluate this integral is by using the Cauchy integral formula:

f(z)=\frac{1}{2\pi i}\oint_C \frac{f(\zeta)}{\zeta-z}d\zeta.
Which holds f is analytic on and inside C and z is any point inside C.

Notice \frac{1}{z+1} is analytic inside (C:|z-i|=1).
This would not be the case for (C: |z|=2).

 

[mathwonk]

the difference between the two circles is in the number of poels inside them, as galileo observed. i.e. the simpelst way to do this integral is by evaluating residues att he poles inside the path. the first path contains one pole, the one at i, and the second path contains two, poles, at i and -i.


the tool is based on local expansions. and cauchy's theorem.

i.e. 1/z is basically dln(|z|) + 2pi idtheta, so when you integrate it around a lop inclosing 0, you get 2pi i.

the same holds if you shift to other points. so 1/(z-i) gives 2pi i when integrated around i, and 1/(z+i) gives 2pi i when intregrated around -i.

each is holomorphic near the other's pole so 1/(z-i) e.g. gives zero when integrated around a small circle centered at -i.