The Cauchy-Goursat Theorem and Contour Integrals

  • Context: Graduate 
  • Thread starter Thread starter Bachelier
  • Start date Start date
  • Tags Tags
    Theorem
Bachelier
Messages
375
Reaction score
0
The Cauchy-Goursat Theorem states:

Let ##f## be holomorphic in a simply connected domain D. If C is a simple closed contour that lies in D, then
##\int_C f(z) \mathrm{d}t = 0 ##​

Now if ##f## is holo just on ##|C| \bigcup \ int(C)## (i.e ##f## holo only on the contour and inside of it, if we take ##z_0 \in \mathbb{C}## can we deduce from the theorem that

## \int_C \frac{f'(z)}{z-z_0} \mathrm{d}t = \int_C \frac{f(z)}{(z-z_0)^2} \mathrm{d}t = 0 ## whether ##z_0 \in \left[\ |C| \bigcup \ int(C) \right]## or not

Since both equal ##0##.

Also where does ##z## must belong to for the theorem to apply?
 
Physics news on Phys.org
I think I got it. I cannot apply Cauchy inside because of the singularity problem.

Now my problem is, even though we don't know if ##f## is holo outside by the givens, can we still apply Cauchy outside ##C## and claim that

[tex]\int_c{f(z)} \mathrm{d}t = 0[/tex]​
 
Last edited:

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K