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Complex integration z^0.5

  1. Oct 2, 2007 #1
    I've been asked to find the value of:

    [tex]\int_{-1}^{1}z^{\frac{1}{2}}dz[/tex]

    Here is how I did it:

    I want to change to polar form. To do that I'll use the fact that:

    [tex]z=e^{i\theta}[/tex] For a unit circle radius 1.
    [tex]dz=ie^{i\theta}d\theta[/tex]

    I replace z and dz:

    [tex]= \int_{-1}^{1}e^{\frac{i\theta}{2}}ie^{i\theta}d\theta[/tex]

    [tex]= \int_{-1}^{1}ie^{\frac{i\theta}{2}}e^{i\theta}d\theta[/tex]

    [tex]= \int_{-1}^{1}ie^{\frac{3i\theta}{2}}d\theta[/tex]

    [tex]= \frac{2}{3}\int_{-1}^{1}\frac{3i}{2}e^{\frac{3i\theta}{2}}d\theta[/tex]

    [tex]= \frac{2}{3}(e^{\frac{3i\theta}{2}})^{-1}_{1}[/tex]

    [tex]= (\frac{2}{3})(e^{\frac{3i}{2}}-e^{\frac{-3i}{2}})[/tex]

    Did I do this correctly? How can I check my work?
     
  2. jcsd
  3. Oct 2, 2007 #2

    Dick

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    Definitely not right. You changed the integration variable from z to theta, but then you plugged the z limits into theta. That's definitely wrong. You also have to be careful about how you define z^(1/2). You know about branch cuts, right?
     
  4. Oct 2, 2007 #3
    Okay thanks. I know a *little* about branch cuts. I think that I need one on the -y axis since I want to trace a circut from theta= 0 to pi. (but not go through the origin.)

    But, how do I change the -1 and 1? I see why what I did was wrong. I didn't change the interval... But, I can't quite see how to change it.
     
  5. Oct 2, 2007 #4

    Hurkyl

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    If z = 1 and z = Exp(i theta)...
     
  6. Oct 2, 2007 #5

    Dick

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    If you know the limits on theta are 0 and pi, those are the limits you should put into your antiderivative, not 1 and -1. Those are the z limits.
     
  7. Oct 2, 2007 #6
    okay

    z=1
    z = Exp(i theta)

    Then theta = 2pi

    z=-1
    z = Exp(i theta)

    then theta = 1 NO!

    then theta = pi !!!

    I don't see how this helps. I think the values of a and b should be 0 and pi, but I don't know WHY.

    I also don't know how to relate the branch cut -pi/4 <= theta < 3pi/4 to all of this.
     
    Last edited: Oct 2, 2007
  8. Oct 2, 2007 #7
    This makes sense to me, but I don't know if I'll know what to do on the next problem, becuse it might not be that easy to see. Is it because I'm changing 1 and -1 from rectangular to polar? And I can just ignore r because it is a constant?
     
    Last edited: Oct 2, 2007
  9. Oct 2, 2007 #8
    Okay lemme try this again. I feel like I'm gonna cry this stuff is so confusing.
    grrr.. Okay

    [tex]z=e^{i\theta}[/tex] For a unit circle radius 1.
    [tex]dz=ie^{i\theta}d\theta[/tex]
    1 becomes 0 and -1 becomes [tex]\pi[/tex] in terms of [tex]\theta[/tex].

    I replace z and dz AND I also replace a and b:

    [tex]= \int_{0}^{\pi}e^{\frac{i\theta}{2}}ie^{i\theta}d\theta[/tex]

    [tex]= \int_{0}^{\pi}ie^{\frac{i\theta}{2}}e^{i\theta}d\theta[/tex]

    [tex]= \int_{0}^{\pi}ie^{\frac{3i\theta}{2}}d\theta[/tex]

    [tex]= \frac{2}{3}\int_{0}^{\pi}\frac{3i}{2}e^{\frac{3i\theta}{2}}d\theta[/tex]

    [tex]= \frac{2}{3}(e^{\frac{3i\theta}{2}})^{\pi}_{o}[/tex]

    [tex]= (\frac{2}{3})(e^{\frac{3i\pi}{2}}-e^{0})[/tex]

    [tex]= (\frac{2}{3})(-i-1)[/tex]

    [tex]= (\frac{-2i-2}{3})[/tex]
    Did I do this correctly? How can I check my work?
     
    Last edited: Oct 2, 2007
  10. Oct 2, 2007 #9

    Dick

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    It's because you are changing variables, period. If I want to integrate f(x^2) from 1 to 2 and I change variables using u=x^2 and I get an antiderivative of g(u), I evaluate that from u=1 to 4. Not 1 to 2.
     
  11. Oct 2, 2007 #10

    Dick

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    I don't want to make you cry, but your integration variables are backwards. I would evaluate it from pi to 0, not 0 to pi. I get (2/3)(1+i). The only way to check that I can think of is to evaluate along another path from -1 to 1 that avoids your branch cut. Wherever it is.
     
  12. Oct 2, 2007 #11
    That makes sense 1^2 = 1 and 2^2=4 and you have u=x^2.

    I'm having a hard time seeing what u is in my problem. Well, I guess it would be theta since it's (d theta) that I'm working with.... so theta = "a function of z"

    z = x + iy

    theta = arctan y/x

    Is that it? arctan 0 ? that makes no sense....
     
    Last edited: Oct 2, 2007
  13. Oct 2, 2007 #12
    No I'm not gonna cry you've been a BIG help! Okay, I mixed up the -1 and the 1. But when I make that change it should change the sign.

    I have 3 more of these to do. Maybe I should go practice some calc 3 change of variable problems before I attempt them. Is that what it's called "change of variables" ? It's been 8 years since I had calc and I'm really rusty.
     
  14. Oct 2, 2007 #13

    Dick

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    You said, "I want to trace a circut from theta= 0 to pi". When did you decide that made no sense?
     
  15. Oct 2, 2007 #14

    Dick

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    Practice is never bad. But I think you know the what the problem is. If you decide to use theta as a variable instead of z, you should use theta limits instead of z limits. That's not rocket science. I'd worry more about branch cut issues.
     
  16. Oct 2, 2007 #15

    It made sense but only becuase I drew a little sketch.

    But I found another error that I made and I think I see how to change a and b in general...

    It's this:

    z=1
    z = Exp(i theta)

    Then theta = 2pi

    z=-1
    z = Exp(i theta)

    then theta = 1 NO!

    then theta = pi !!!
     
  17. Oct 2, 2007 #16

    Dick

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    Sketches also help. Now start drawing branch cuts. They are pretty important with functions like z^(1/2).
     
  18. Oct 3, 2007 #17

    HallsofIvy

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    One problem I see right from the start is that you do not specify any path for the integral. If z is a complex variable then there are an infinite number of paths you could use to go from -1 to 1. Taking [itex]z= e^{i\theta}[/itex] with [itex]\theta[/itex] from [itex]-\pi[/itex] to [itex]\pi[/itex] is choosing the upper unit semi-circle. Seeing only "-1 to 1" my first thought would have been to use the x-axis as path. Are those two integrals the same?
     
  19. Oct 3, 2007 #18
    I don't think you can go through the origin. It's a branch point. That means you can't take that path or integral. (Right?)

    I'm still fuzzy on how to use the branch cut. I know it needs to go from the origin to [tex]z_{\infty}[/tex], and if I think of the stereographic projection that means that any line from the origin off in some direction will do. So, I'll make it go down along the -y axis... so it won't hit my line integral in the upper half plane. You can't take the intergral and go through a branch cut.

    What impact does this choice have on my answer? Or is the branch cut only important for cases like:

    [tex]\int_{C}^{}\frac{dz}{z}[/tex] on a closed circuit called "C" around the origin, where if you ignore the branch cut you might think the value is 0 (and this is not correct.)
     
  20. Oct 3, 2007 #19

    Dick

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    You CAN go through the origin, if you are careful, you just don't want to go through the negative y axis. The values of the function from 0 to 1 are just x^(1/2) and the values from -1 to 0 are i*|x|^(1/2). So you can see that the absolute values you are integrating are the same over both intervals and the absolute value of each integral is 2/3. So the [-1,0] gives i*2/3 and [0,1] gives 2/3. Total (2/3)*(1+i).
     
  21. Oct 3, 2007 #20
    Wow. I didn't know that. Thank you so much for you help!
     
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