# Why the path independence theorem does not work?

1. Nov 7, 2007

### oahsen

1. The problem statement, all variables and given/known data
where C is the contour given with direction marked by increasing y, and where -2≤y≤2 , compute itgeral(z^2-2z+1)dz. With the condition x=5;

Firstly I solved the auestion with the classical way ;
taking z= 5 + it where -2≤t≤2;
we take the i*integral((5+it)^2-2(5+it) +1)dt from t=-2 to t=2 and I found the result as
176i/3.

Then I tried to use the path independence theorem. The antiderivative of z^2-2z+1 is z^3/3 - z^2 + z then taking the integral from 2i to -2i yields -4i/3. So the two answers are different. Why the path independence theorem did not work here?[/QUOTE]

Last edited: Nov 7, 2007
2. Nov 7, 2007

### Hurkyl

Staff Emeritus
That makes no sense; you have to integrate with respect to some variable. It looks like you just erased the dz from the original integral before getting to this point!

3. Nov 7, 2007

### oahsen

sorry , I made a mistyping now I corrected it. There I used the formula
integral(f(z)dz) over a contour c = integral(f(z(t))* z'(t)) dt .

4. Nov 7, 2007

### Hurkyl

Staff Emeritus
Your paths don't have the same endpoints, do they?

Incidentally,
It's not the antiderivative; it's an antiderivative. (But that's not relevant here)