# Cauchy Integral Extension Complex Integrals

1. Oct 15, 2009

### ryanj123

1. The problem statement, all variables and given/known data
Allow D to be the circle lz+1l=1, counterclockwise. For all positive n, compute the contour integral.

2. Relevant equations

int (z-1/z+1)^n dz

3. The attempt at a solution

I know to use the extension of the CIF.

Where int f(z)/(z-zo)^n+1 dz = 2(pi)i* (f^(n)(zo)/n!) ....

However, I'm unsure how to execute the integral for my answer to depend on n.

Then,

Int ( f(z)/(z+1)^n) = 2(pi)i*(f^n(zo)/n!)

Evaluating

f(zo) at zo=-1

= (-2)^n

So..

2(pi)i*((-2)^n/n!)

I know I'm missing components of the derivative operator... but I'm not sure how to go about completing this.

I appreciate any help.

2. Oct 15, 2009

### Donaldos

$$f^{(n)}(z_0)=\left.\frac{{\rm d}^n f}{{\rm d} z^n}\right|_{z=z_0}\neq \left(f(z_0)\right)^n$$

3. Oct 15, 2009

### ryanj123

I'm trying to go back...

Int ((z-1)/(z+1))^n dz

If f(z) = (z-1)^(n)

Then,
Int (f(z)/(z+1)^n)

Where zo=-1

So,
2(pi)i*f^(n)(-1)/n!

For any n>0

Is this sufficient to assume? Then for whichever n is used, f(z) can be differentiated the amount of times and evaluated at (-1) as needed.

4. Oct 15, 2009

### soopo

Good question!

Your problem seems to be in this part

$$\int = 2\pi i Res ( \frac {d^{n-1} } {dz^{n-1}} (z - 1)/n! )$$
// I am not completely sure about how to write the Residue part

So what is the derivate of (z-1)?
The second derivate?
...
The n'th derivate?