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

Complex integral question

  • Thread starter Applejacks
  • Start date
  • #1
33
0

Homework Statement




Assuming a counterclockwise orientation for the unit circle, calculate
∫ [itex]\frac{z+i}{z^3+2z^2}[/itex] dz
|z|=1

Homework Equations



f'(a)=[itex]\frac{n!}{2i\pi}[/itex]=∫[itex]\frac{f(z)}{{z-a}^(n+1)}[/itex]
???

The Attempt at a Solution



I don't understand these types of questions. What does the |z| have to do with the integral? It's written on the bottom limit of the integral in case that wasn't clear.

From the answers, it writes:
f(z)=[itex]\frac{z+i}{z+2}[/itex]
2[itex]\pi[/itex]f'(0)=[itex]\frac{\pi}{2}[/itex]+[itex]\pi[/itex]i


Edit: Nevermind. I found out how the formula works. I still don't understand what the |z| does. What would happen if it was |z|=10 for example?
 
Last edited:

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
|z|=1 is represents the closed curve (unit circle) around which the contour integration should be performed.

do you know the residue theorem, could be pretty useful here
 
  • #3
33
0
I somewhat understand the residue theorem and how singularities work. I'm not sure how that explains what the circle does. As long as the singularity is located in the circle, then I can do the integral. If I had something like |z-2|=1, then a problem would occur since the singularity is outside the circle. How do I adjust it then?
 
  • #4
lanedance
Homework Helper
3,304
2
you don't the residue theorem relates the integral around a closed curve to the poles located inside the curve - so you need to find which poles are located with the unit circle

start with finding the poles of the function
 
  • #5
33
0
So if I had something like the expression below, where C is the square with vertices ([itex]\pm2+2i[/itex],[itex]\pm2-2i[/itex])

∫ [itex]\frac{cos(z)}{z(z^2+8)}[/itex] dz
C

Would I ignore the poles at +/-isqrt(8) and proceed to use the CIF?

2[itex]\pi[/itex]i*f(0)=2[itex]\pi[/itex]i*[itex]\frac{cos(0)}{(0^2+8)}[/itex]=[itex]\frac{i\pi}{4}[/itex]
 
Last edited:
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,187
Yup, you only care about the poles inside the contour.

To answer your question in post 3: If there are no singularities inside the contour, the integral is equal to 0.
 
Last edited:
  • #7
33
0
To answer your question in post 3: If there are no singularities inside the contour, the integral is equal to 0.
omg I feel so stupid for not seeing something this simple...

Thanks for the help so far. I have one last question. What happens when you have multiple singularities within the contour. For example:

∫ [itex]\frac{sin(z)cos(z)}{((z-\frac{\pi}{4}}[/itex])(z+[itex]\frac{\pi}{4}[/itex]) dz
|z|=[itex]\pi[/itex]


Edit: blah I can't fix the fractions in the integral. The z+pi/4 is on the denominator

Do we treat each one individually and then add them up?

2[itex]\pi[/itex]i*f([itex]\frac{\pi}{4}[/itex])=2[itex]\pi[/itex]i*[itex]\frac{1}{2*(\pi/2}[/itex])=2i

2[itex]\pi[/itex]i*f([itex]\frac{-\pi}{4}[/itex])=2[itex]\pi[/itex]i*[itex]\frac{-1}{2*(-\pi/2)}[/itex]=2i

2i+2i=4i
 
Last edited:
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,187
Yes, that's what you do. The integral is equal to [itex]2\pi i\sum~\mathrm{residues}[/itex].
 
  • #9
33
0
Awesome. I finally understand this topic! Thanks Vela and Lanedance.
 

Related Threads for: Complex integral question

Replies
6
Views
3K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
0
Views
666
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
598
Replies
1
Views
741
Replies
1
Views
1K
Replies
10
Views
4K
Top