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Math Help - so confused

  1. #1
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    so confused

    Using the parametric representation
    z(t) = e^(it)= cost + isint how do we evaluate the following integral
    (a) integral from 1 to -1 (1/z)dz along |z|=1, the upper half plane.


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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sandy View Post
    Using the parametric representation
    z(t) = e^(it)= cost + isint how do we evaluate the following integral
    (a) integral from 1 to -1 (1/z)dz along |z|=1, the upper half plane.


    I'm not quite sure what you're asking. Is it to evaluated \int_C \frac{dz}{z} with C=\left\{e^{it}:t\in[0,\pi]\right\}?
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  3. #3
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    If z= e^{i\theta}, then [tex]\frac{1}{z}= z^{-1}= e^{-i\theta}, of course.

    As \theta goes from 0 to [itex]\pi[/itex], z with go over the upper unit semicircle so z covers C.

    \int_C \frac{1}{z} dz= \int_0^\pi e^{-i\theta} d\theta

    Since it is also true that e^{i\theta}= cos(\theta)+ i sin(\theta) you can do that integral in terms of sin and cosine.
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    I'm not quite sure what you're asking. Is it to evaluated \int_C \frac{dz}{z} with C=\left\{e^{it}:t\in[0,\pi]\right\}?
    Yes thats reight im asking that and the integral is from -1 to 1
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    If z= e^{i\theta}, then [tex]\frac{1}{z}= z^{-1}= e^{-i\theta}, of course.

    As \theta goes from 0 to [itex]\pi[/itex], z with go over the upper unit semicircle so z covers C.

    \int_C \frac{1}{z} dz= \int_0^\pi e^{-i\theta} d\theta

    Since it is also true that e^{i\theta}= cos(\theta)+ i sin(\theta) you can do that integral in terms of sin and cosine.
    is the answer -2i ??
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  6. #6
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    Quote Originally Posted by sandy View Post
    is the answer -2i ??
    can someone please check if this is right
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