• Support PF! Buy your school textbooks, materials and every day products Here!

Complex Variables

Can anyone help me with this pls?

How can you prove that the integral of f(z) around the contour z= 1 is 0

where f(z) is Log(z+5)

Thx





I know Log(z) is ln r + i (theta). But i dont know how that applies to this situation.

Also, do I solve it as a normal integral or use the Cauchy Goursat theorem to prove that its integral is zero?
 

Answers and Replies

tiny-tim
Science Advisor
Homework Helper
25,789
249
Welcome to PF!

How can you prove that the integral of f(z) around the contour z= 1 is 0

Also, do I solve it as a normal integral or use the Cauchy Goursat theorem to prove that its integral is zero?
Hi soulsearching! Welcome to PF! :smile:

Do you mean |z| = 1?

If so, then z + 5 is just the circle of radius 1 and centre at 5. It doesn't include the origin (z + 5 = 0), so yes … always do it the easy way … use Cauchy Goursat! :smile:
 
Hi soulsearching! Welcome to PF! :smile:

Do you mean |z| = 1?

If so, then z = 5 is just the circle of radius 1 and centre at 5. It doesn't include the origin (z + 5 = 0), so yes … always do it the easy way … use Cauchy Goursat! :smile:


So the point z = 5 is outside the contour, so the integral vanishes right? Thank you Tim.



When trying to show the same for this equation here around the same contour

f(Z) = (4z^2 -4z +5)^-1

I found the singularities here to be (1/2 + i) and (1/2-i). Is the integral zero because both singularities are outside the circle of radius 1? Thank you.

I have a major exam tmr and my professor is not being of much help, but I really appreciate your help. :-)
 
229
0
So the point z = 5 is outside the contour, so the integral vanishes right? Thank you Tim.



When trying to show the same for this equation here around the same contour

f(Z) = (4z^2 -4z +5)^-1

I found the singularities here to be (1/2 + i) and (1/2-i). Is the integral zero because both singularities are outside the circle of radius 1? Thank you.

I have a major exam tmr and my professor is not being of much help, but I really appreciate your help. :-)
Yes.

And in this case your integrand is analytic inside and on the circle |z| = 1, so again it's 0 by cauchy's.
 
Thanks Dan.
 
tiny-tim
Science Advisor
Homework Helper
25,789
249
When trying to show the same for this equation here around the same contour

f(Z) = (4z^2 -4z +5)^-1

I found the singularities here to be (1/2 + i) and (1/2-i). Is the integral zero because both singularities are outside the circle of radius 1?
Looks good to me! :smile:
 
Also when trying to find the integral of (1/8z^3 -1) around the contour c=1.

I found the singularities to be 1/2, 1/2exp(2pi/3), and 1/2exp(4pi/3)

What is the next step here. Do I just assume the integral is 6pi(i) after using partial fractions to find the numerators of the 3 fractions.

The answer is actually 0, but I dont understand how it was solved. I know it might have something to do with the rule that says, the sum of the integral of a C1, C2 and C3 is equal to the integral of the outside contour, but how do i know the orientation of the three contours inside the big contour C.

Thanks, hope im not asking too many questions
 

Related Threads for: Complex Variables

  • Last Post
Replies
2
Views
705
  • Last Post
Replies
0
Views
982
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
692
  • Last Post
Replies
1
Views
818
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
Top