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Basic Complex Integrals

  1. Jun 22, 2011 #1
    I'm integrating 1/(z-1/2) over the closed disk w/ radius = 3 centered at 0.

    I've seen other problems where the final answer was i2pi times f(w) - here w =1/2.

    Since f(z) is equal to 1. Is the final answer just i2pi?

    Next up:

    I have the integral of dt/(2 + sint) the problem then tells me expand sin to its complex definition and replace e^it with z. My question is what does dt become?


    Lastly:

    I'm given the integral zexp(z^2). I'm asked to provide a proof for why it equals zero on any closed curve gamma.

    Do I simply perform a u sub (u=z^2, du=2zdz) and then say that since e^u is entire and thus holomorphic on any region (or curve) it is equal to 0?
     
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  3. Jun 22, 2011 #2

    micromass

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    Hi MurraySt! :smile:

    The answer to your first question is indeed [itex]2\pi i[/itex], but you might want to give a more rigourous reason for this. Maybe the Cauchy Integral formula.

    For the second question, consider [itex]\gamma[/itex] the circle of radius 1, thus [itex]\gamma(t)=e^{it}[/itex] You must find a complex function such that

    [tex]\int_\gamma{f(z)dz}=\int_0^{2\pi}{\frac{dt}{2+\sin(t)}}[/tex]

    (I just guessed the limits, as you did not provide any). So using the definition of the complex integral, you must find f such that

    [tex]\int_0^{2\pi}{f(e^{it})ie^{it}dt} = \int_0^{2\pi}{\frac{dt}{2+\sin(t)}}[/tex]

    Regarding the third question. Can't you just say that [itex]ze^{z^2}[/itex] is holomorphic on [itex]\mathbb{C}[/itex] and thus any integral is 0.
     
  4. Jun 23, 2011 #3
    As a quick follow-up: If my curve is oriented negatively (clockwise) how does this affect my final answer? Do I simple negate it?
     
  5. Jun 23, 2011 #4

    micromass

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    Yes! :smile:
     
  6. Jun 23, 2011 #5
    So sorry to bother you one final question:

    The denominator of my problem is z^2 +4iz -1. I used the quadratic equation to get the answers i(-2+rad3) and i(-2-rad3).

    Does my denominator simply become the product of the roots? It scares me that I don't get back any z.

    Thanks as always for your help!
     
  7. Jun 23, 2011 #6

    micromass

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    Well, you're scared for the right reason!! In general, if we have

    [tex]az^2+bz+c[/tex]

    and if we have roots [itex]\xi_1,\xi_2[/itex], then we can factor it as

    [tex]az^2+bz+c=a(z-\xi_1)(z-\xi_2)[/tex]
     
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