Cauchy Integral Extension Complex Integrals

[ryanj123]

Homework Statement

Allow D to be the circle lz+1l=1, counterclockwise. For all positive n, compute the contour integral.

Homework Equations

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

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.

I made f(z)=(z-1)^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.

 

[Donaldos]

[tex]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[/tex]

 

[ryanj123]

[tex]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[/tex]

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.

 

[soopo]

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.

Good question!

Your problem seems to be in this part

[tex] \int = 2\pi i Res ( \frac {d^{n-1} } {dz^{n-1}} (z - 1)/n! ) [/tex]
// 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?