Contour Integral Homework Statement Evaluation

elimenohpee
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Homework Statement


Let C be a contour formed by the points O(0,0), A(1,0), B(1,1), with the direction OA->AB->BO. By using the definition of a contour integral, evaluate:

(integral) f(z)dz

Homework Equations



\int f[z(t)]z'(t)dt

The Attempt at a Solution


I didn't include the work I've done, or even the function upon which I am integrating. I'm looking more for an understanding.

I perform the line integrals about each side of the triangle, and sum each value at the end. But I end up with 0, which I think makes sense since this is a closed region. But should the value equal zero?
 
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It might be zero. But since you skipped telling us what the function f(z) is, it's hard to say. It also might not be zero. Is that what you wanted to know? What is f(z)?
 
f(z) = e^pi*z

Its a little too much to try and type all the work I did computing each integral.
 
elimenohpee said:
f(z) = e^pi*z

Its a little too much to try and type all the work I did computing each integral.

That's fine. Since e^(pi*z) doesn't have any poles, then you are ok. The integral should come out to be zero. It's analytic inside the triangle. That's Cauchy's integral theorem. There are other functions that don't satisfy that criterion.
 
Dick said:
That's fine. Since e^(pi*z) doesn't have any poles, then you are ok. The integral should come out to be zero. It's analytic inside the triangle. That's Cauchy's integral theorem.

Thank you, exactly what I figured and actually worked out. Just wanted to verify.

Thanks again!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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