Contour Integration

  • #1

Homework Statement




∫[itex]_{\gamma}[/itex](x-y)dz where [itex]\gamma[/itex] has parametrization: z(t) = e[itex]^{it}[/itex] for [itex]\pi[/itex]/2 [itex]\leq[/itex] t [itex]\leq[/itex] 3[itex]\pi[/itex]/2

Homework Equations


the integral of the sum is the sum of the integral


The Attempt at a Solution


I tried to break it up and see if I could evaluate it as I normally would but it started to get really messy and I think I was going about it wrong.

z=exp(it) and dz=iexp(it)
∫(x)dz - i∫(y)dz

can someone please push me into the right direction? thank you
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
as you mean to evaluate directly you need to use a substitute for the parameterisation variable

[tex]
\cint_{\gamma} f(z) dz = \int_{t_a}^T_b f(z(t)) z'(t) dt[/tex]
 
  • #3
could you elaborate just a little more? I really appreciate your help
 
  • #4
lanedance
Homework Helper
3,304
2
what don't you understand? I won't do the problem for you, but am happy to help
 
  • #5
I dont understand how to set up a substitution so I can get it into a form that I know how to work with
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,847
966
You intend that z= x+ iy, right? So if [itex]z= e^{it}= cos(t)+ i sin(t)[/itex] what are x and y in terms of t?
 
  • #7
Okay this is what I have so far:

∫[itex]_{\gamma}[/itex](x-y)dz where [itex]\gamma[/itex] has parametrization: z(t)=e[itex]^{it}[/itex] for [itex]\pi[/itex]/2[itex]\leq[/itex]t[itex]\leq[/itex]3[itex]\pi[/itex]/2


∫[itex]_{\gamma}[/itex](x-y)dz = ∫[itex]_{\gamma}[/itex](cos(t)-isin(t))dz

which we can break up like,

∫[itex]_{\gamma}[/itex](cos(t))dz - i∫[itex]_{\gamma}[/itex](sin(t))dz

since z(t)=e[itex]^{it}[/itex] then dz=ie[itex]^{it}[/itex]

so we have

∫[itex]_{\gamma}[/itex](cos(t))(ie[itex]^{it}[/itex]) - i∫[itex]_{\gamma}[/itex](sin(t))(ie[itex]^{it}[/itex] )

is this correct so far?

or would my dz be -sin(t)+icos(t)?
 
Last edited:
  • #8
lanedance
Homework Helper
3,304
2
you may want to check your "y" value, remember z= x+ iy,
 
  • #9
so instead of isin(t) just simply sin(t)

∫[itex]_{\gamma}[/itex](x-y)dz = ∫[itex]_{\gamma}[/itex](cos(t)-sin(t))dz

∫cos(t)dz-∫sin(t)dz

every time I try evaluating the above expression I never can seem to get the correct answer. I'm using ie^(it) for my dz

AHHH im getting so frustrated. I feel like I am missing an important concept, this problem should not take me this much time
 

Related Threads on Contour Integration

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
2
Views
815
  • Last Post
Replies
10
Views
761
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
10
Views
2K
Top