1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Contour integrals

  1. Aug 13, 2012 #1
    1. The problem statement, all variables and given/known data

    calculate the contour integral [itex]\oint_{C} (y^2+ix)dz[/itex] where C consists of the parabolic path z(t)=t[itex]^{2}[/itex]+it for 0≤t≤1 followed by the straight line segment from the point 1+i to the point 0
    2. Relevant equations



    3. The attempt at a solution

    so the contour is in 2 parts for the first part [itex]\oint_{C} (y^2+ix)dz[/itex] = [itex]\int^{1}_{0} (t^2 + it)(2t+i)dt = \int^{1}_{0} (2t^3 + i3t^2-t)dt = i [/itex]

    and second part is integral of the straight line from 1+i to 0. this can be represented by z(t) = t+ti for 1≤t≤0

    [itex]\int^{0}_{1} (t+ti)(1+i)dt = \int^{0}_{1} (2ti)dt = -i[/itex]

    and when i add them together i get 0. is this right? it asks what can i deduce about the function y[itex]^{2}[/itex]+ix ?
     
  2. jcsd
  3. Aug 13, 2012 #2
    Why do you write the integrals that way? What is the general formula?
     
  4. Aug 13, 2012 #3
    hi voko,
    Not sure what you mean but what i did was
    step 1 represent curve C as z(t) in terms of t. (in this case it was given)
    step 2 represent the function f(z) as f[z(t)] in terms of t
    step 3 take the derivative of z(t) in terms of t [itex]\frac{dz}{dt}[/itex]
    step 4 sub all the above into [itex]\int^{b}_{a} f[z(t)] \frac{dz}{dt} dt[/itex]
     
  5. Aug 13, 2012 #4
    Step 2 is not done correctly in both cases. You just take z(t) instead of f(z(t)).
     
  6. Aug 13, 2012 #5
    in the question i took z(t) = t[itex]^{2}[/itex] + it and [itex]\frac{dz}{dt} = 2t + i[/itex] ans subed them in is this not right?
     
  7. Aug 13, 2012 #6
    Remember, z = x + iy. f(z) = y2 + ix. What does f(z) look like when z = x + iy = t2 + it?
     
  8. Aug 13, 2012 #7
    dohhhh

    I see now it should be f(z) = t[itex]^{2} + it^2[/itex]

    thanks a million voko
     
  9. Aug 15, 2012 #8
    just so im doing this right my integral then becomes [itex]\int^{1}_{0} (t^2 + it^2)(2t+i2t)dt = \int^{1}_{0} (2t^3 + i2t^3+i2t^3-2t^3)dt = \int^{1}_{0} (i4t^3)dt = i [/itex]
     
  10. Aug 15, 2012 #9
    No, this does not look right!

    [tex]\int^{1}_{0} (t^2 + it^2)(2t+i2t)dt = \int^{1}_{0} t^2(1 + i)2t(1 + i)dt = \int^{1}_{0} 2t^3(1 + i)^2dt = (1 + i)^2 \int^{1}_{0} 2t^3 dt \ne \int^{1}_{0} (i4t^3)dt[/tex]

    And don't forget about the second part of the path.
     
  11. Aug 15, 2012 #10
    i dont understand they are equal [itex] (1 + i)^2 \int^{1}_{0} 2t^3 dt = \int^{1}_{0} (i4t^3)dt [/itex]
     
  12. Aug 15, 2012 #11
    [itex] (1 + i)^2 = 2i [/itex]
     
  13. Aug 15, 2012 #12
    Indeed :) Somehow I was looking at (1 + i) as if it were (i - i). Sorry about that.

    What about the second part?
     
  14. Aug 15, 2012 #13
    i havent done it out yet but when i do its going to be i as its around a simple closed curve so the have to be the same as one integral minus the other =0 am i right in my thinking on this?
     
  15. Aug 15, 2012 #14
    An integral is zero over a closed curve only if the integrand is holomorphic. So do the second part, get the result and see what it means regarding this.
     
  16. Aug 16, 2012 #15
    [itex] \int^{1}_{0} (t^2 + it)(2t+i)dt [/itex]= [itex] \int^{1}_{0} 2t^3 + i3t^2 - t dt [/itex] = [itex]( \frac{1}{2} t^4 + i t^3 - \frac{1}{2} )^{1}_{0} = i[/itex] thus showing that integrand in anaylitic and holomorphic. The integrand doesnt depend on the path?
     
  17. Aug 16, 2012 #16
    How did you get (2t + i)?
     
  18. Aug 16, 2012 #17
    f'[z(t)]?
     
  19. Aug 16, 2012 #18
    or should it be [itex] \int^{1}_{0} (t^2 + it)(1+i)dt = \int^{1}_{0} t^2 + it^2 +it -t dt = \frac{1}{3} + i\frac{1}{3} -i\frac{1}{2} - \frac{1}{2} [/itex]
     
  20. Aug 16, 2012 #19
    dz in the original integral integral is replaced with z'(t)dt, which gives (1 + i)dt as you have it here correctly. So, what can be said about the holomorphicity of the integrand? Care to check the Cauchy-Riemann equation?
     
  21. Aug 16, 2012 #20
    i also had the first part wrong it should be [itex]\int^{1}_{0} (t^2 + it^2)(2t+i)dt = \int^{1}_{0} (2t^3 + it^2+i2t^3-t^2)dt = \frac{1}{2} + i\frac{1}{3} + i\frac{1}{2} - \frac{1}{3} [/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Contour integrals
  1. Contour integrals (Replies: 4)

  2. Contour integral (Replies: 3)

  3. Contour integral (Replies: 1)

  4. Contour integration (Replies: 4)

  5. Contour integrals (Replies: 2)

Loading...