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Homework Help: Complex Analysis Integral

  1. Oct 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Evaluate [tex]\oint_C \f(z) \, dz [/tex] where C is the unit circle at the origin, and f(z) is given by the following:

    A. [tex]e^{z}^{2}[/tex]
    (the z2 is suppose to be z squared)
    B. [tex]1/(z^{2}-4)[/tex]

    2. Relevant equations



    3. The attempt at a solution
    I'm completely confused
     
    Last edited: Oct 10, 2007
  2. jcsd
  3. Oct 10, 2007 #2

    malawi_glenn

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    okay

    You know this theorem:

    if f(z) is analytic inside and on a closed loop C; then the integral of f(z) around the loop is zero

    is [tex] e^{z^2} [/tex] analytic inside and on the unit circle?

    b. write z^2 - 4 as (z-2)(z+2); is [tex]1/(z^{2}-4)[/tex] analytic inside and on the unit circle?
     
  4. Oct 11, 2007 #3
    Okay, so for

    a. I get that [tex] e^{z^2} [/tex] coincides with the unit circle at 1. But isn't it analytic at 1? Shouldn't the integral be zero? I'm not quite sure where to go from here.

    b. I get that the integral is zero because the singularities are outside of the unit circle.
     
  5. Oct 11, 2007 #4

    malawi_glenn

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    [tex] e^{z^2} [/tex] is analytic on and inside the unit circle.

    Infact [tex] e^{z^2} [/tex] is entire, i.e analytic in the whole complex plane.

    You should not calculate what the funtions are at certain values of z, you should just find out where they are analytic and where they are not. Then you use Cauchys integral theorem and other theorems (in the case when you have closed loops).
     
  6. Oct 11, 2007 #5
    Okay, so I have a similar problem


    [tex]\oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz [/tex]

    Here the singularities are z=0, and z= [tex]\Pi[/tex]
    Find the integral where
    a. C is the boundary of the annulus between |z|=1 and |z|=4. So, this C encloses z=[tex]\Pi[/tex]
    b. C is the circle of radius R where R<[tex]\Pi[/tex]. So, this C encloses z=0.
     
    Last edited: Oct 11, 2007
  7. Oct 11, 2007 #6

    malawi_glenn

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    \pi

    Gives you the "pi" in tex =)

    And use \dfrac{nominator}{denominator}

    To make nice fractions

    Just so we can see exaclty how the integral looks like.

    use this:
    http://en.wikipedia.org/wiki/Cauchy's_integral_formula
     
    Last edited: Oct 11, 2007
  8. Oct 11, 2007 #7
    Okay I think that's about as close as I'm going to get lol
     
    Last edited: Oct 11, 2007
  9. Oct 11, 2007 #8

    malawi_glenn

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    no just \pi

    \Pi gives you the big pi =)

    The trick is to for each case to identify the f(z) and the "a", then you just take the derivative of f(z) and evaluate it at z=a; then multiply with [tex] 2 \pi i [/tex] in this case, since that factor is excluted in the problem, but must be there to use the formula.

    Do you have a textbook on complex analysis?
     
  10. Oct 11, 2007 #9
    Yes I do - complex variables by fokas and ablowitz, but I dont find it very helpful.
     
  11. Oct 11, 2007 #10

    malawi_glenn

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    okay, Saff and Snider is awesome I think.

    But now, what do you think about the integral, can you give it a try and if you get stuck post exaclty what you did and thought here so we can help you.
     
  12. Oct 11, 2007 #11
    Okay.
    The one I've tried so far is the one where C is the circle of radius R such that R> [tex]\pi[/tex]. So, this circle would contain both singularities.

    So,
    [tex]\oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz [/tex]

    = [tex]\oint_C \frac{\frac{e^{iz}}{z}}{(z-\pi)}\, dz [/tex] + [tex]\oint_C \ \frac{\frac{e^{iz}}{z-\pi}}{z}\, dz [/tex]

    = [tex]2i\pi \ (\frac{e^{iz}}{z}) + 2i\pi \ (\frac{e^{iz}}{z-\pi}) = 2i\pi \ (\frac{e^{iz}}{\pi} + \frac{e^{iz}}{-\pi}) = -4i[/tex]
     
    Last edited: Oct 11, 2007
  13. Oct 11, 2007 #12
    For the one where C is the boundary of the annulus between |z|=1 and |z|=4, and therefore encloses [tex]\pi[/tex], I get:


    [tex]\oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz [/tex]
    =
    [tex]\oint_C \frac{\frac{e^{iz}}{z}}{(z-\pi)}\, dz [/tex]
    = [tex]2i\pi (\frac{e^{iz}}{z} = 2ie^{i \pi} = -2i[/tex]

    (Ugh, I feel like I'm doing something very wrong)
     
    Last edited: Oct 11, 2007
  14. Oct 11, 2007 #13
    Finally, for the one where C is the circle of radius R such that R<[tex]\pi[/tex], I get:


    [tex]\oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz [/tex]

    = [tex]\oint_C \frac{\frac{e^{iz}}{z-\pi}}{(z)}\, dz [/tex]

    = [tex]2i\pi (\frac{e^{iz}}{z-\pi} = 2i\pi \frac{e^{i\pi}}{-\pi}

    = 2i[/tex]

    I am not sure if I am the only who cannot see the final solution to this problem, but in the end i get 2i
     
    Last edited: Oct 11, 2007
  15. Oct 11, 2007 #14
    i think the one for the circle of radius greater than pi is right...
     
  16. Oct 11, 2007 #15

    malawi_glenn

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    According to the problem you told me in post #5 you had only two loops. One that encloses [tex] \pi [/tex] and one that encloses 0.

    a.
    [tex]\oint_{C1} \frac{e^{iz}}{z(z-\pi)}\, dz [/tex]

    let [tex] C1 [/tex] be curve that encloses [tex] \pi [/tex] but not 0.

    [tex]\oint_{C1}\frac{\frac{e^{iz}}{z}}{(z-\pi)}\, dz = 2 \pi i \frac{e^{i \pi}}{\pi} = -2i [/tex]

    b.
    [tex]\oint_{C2} \frac{\frac{e^{iz}}{z-\pi}}{(z)}\, dz [/tex]


    let [tex] C2 [/tex] be curve that encloses 0 but not [tex] \pi [/tex].

    [tex]\oint_{C2} \frac{\frac{e^{iz}}{z-\pi}}{(z)}\, dz = 2 \pi i \dfrac{e^{0}}{0-\pi} = -2i[/tex]

    If you had a big loop that encloses both singularites; then you use the decompisition theorem of loops, and simply add upp the result from the two separate integrals =)

    i.e -2i + (-2i) = -4i

    Also the direction of the curve matters, if you go positive or negative.
     
    Last edited: Oct 11, 2007
  17. Oct 11, 2007 #16
    I almost had it! Thank you so much for your help.
    Also, does that mean that post number 11 is right?
     
  18. Oct 11, 2007 #17
     
    Last edited by a moderator: May 3, 2017
  19. Oct 11, 2007 #18
    Really? I used it for all of them - does that mean that my homework is completely wrong?
     
    Last edited: Oct 11, 2007
  20. Oct 11, 2007 #19
    I have no idea how to do it using the Taylor series
     
  21. Oct 11, 2007 #20
    I don't know, to be honest. Initially I tried substitution, but I ended up with something impossible to integrate using normal techniques. You can integrate some difficult expressions using taylor series and he mentioned that in class, and it's mentioned in the problem where it references the part of the book with all of the taylor series.

    The thing is I can't find an example anywhere of a problem that uses taylor series for something as complex as this. I'm also fuzzy on how or where the series needs to be "centered" --

    As the hour is growing late I worked out my answers using the formula, but I'm still trying to find a way to do it without the formula.
     
  22. Oct 11, 2007 #21
    Well, hopefully he will accept what we did - and maybe he'll be proud that we read ahead (or maybe not lol). But I don't know how to do it the other way and I do remember him mentioning using either the taylor series and the power series - integrating term by term but - since I don't know how to do it I had no choice but to do it this way
     
  23. Oct 11, 2007 #22

    malawi_glenn

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    futurebird; this is complex analysis. And they way to solve these integrals is by using Cauchys integral theorem. This is just basic integration in complex analysis.


    milky: You did not take use of cachys integral formula in your post #11; you got the right result but not in the correct way.

    I think that Cauchys integral formula is one of the coolest results in complex analysis, and also the theory of residue's that is a combination of cauchy integral formula and Laurent series.

    If you wanted to solve these integrals by Taylor expansions and so on, then it should be adressed in the first post in the template "Know formulas" or whatever it is called again, so you could get proper help.
     
    Last edited: Oct 11, 2007
  24. Oct 11, 2007 #23
    Can we address this anyway? I'm very curious.

    I assumed #11 was using the integral formula. Otherwise how would you do that?

    Let me type up what I've done.
     
  25. Oct 11, 2007 #24

    malawi_glenn

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    When you start a new thread in one of the "Homework" forums; this templete will come:

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    So therefore, try using that 2. Writing what relations you know of and so on (not all, but some)

    Now Futurebird, are you and Milky buddies?

    And yes, post #11 he tried using Cauchy integral formula but did some errors.

    here you should have:

    C is a loop containing both pi and zero. and is positive oriented.
    Ca is the loop containg just pi and Cb contain only 0; this we can do by the decomposition of loop/curves theorem. Ca and Cb are positive oriented.

    [tex]\oint_C \frac{e^{iz}}{z(z-\Pi)}\, dz = [/tex]

    [tex]\oint_{Ca} \frac{\frac{e^{iz}}{z}}{(z-\pi)}\, dz + \oint_{Cb} \ \frac{\frac{e^{iz}}{z-\pi}}{z}\, dz =
    [/tex] Now identify the different f(z)'s and evaluate them at z = 0 and z= pi, then multiply the whole thing with 2 pi i :

    [tex]2i\pi \ (\frac{e^{i \pi}}{\pi}) + 2i\pi \ (\frac{e^{i*0}}{0- \pi}) = -4i[/tex]
     
  26. Oct 11, 2007 #25
    [tex]\oint_{C}\frac{e^{iz}}{z(z-\pi)}dz[/tex]

    C is the circle around z =0

    Let
    [tex]z=re^{i\theta}[/tex] and
    [tex]dz=ire^{i\theta}d\theta[/tex]

    subbing that in:

    [tex]\int^{2\pi}_{0}\frac{e^{rie^{i\theta}}ire^{i\theta}}{re^{i\theta}(re^{i\theta}-\pi)}d\theta[/tex]

    [tex]\int^{2\pi}_{0}\frac{ie^{rie^{i\theta}}}{re^{i\theta}-\pi}d\theta[/tex]

    Now I tried using this:

    [tex]e^{z}=\displaystyle\sum_{j=0}^{\infty}\frac{z^{j}}{j!}[/tex]

    to replace:

    [tex]e^{rie^{i\theta}}[/tex]

    But it looks like too much of a mess to work with term by term.

    [tex]\int^{2\pi}_{0}\frac{1+ire^{i\theta}+\frac{-r^{2}e^{2i\theta}}{2!}+\frac{-ir^{3}e^{3i\theta}}{3!}+...}{(re^{i\theta}-\pi)}d\theta[/tex]
     
    Last edited: Oct 11, 2007
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