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Math Help - Integration in a complex plane

  1. #1
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    Integration in a complex plane

    Hello,

    I have a problem:

    Let gamma: z=e^(it) (0<= t <= pi). Show that

    |integral (e^z / z) dz | <= pi e^2

    I thought to use the Estimation theorem:

    But at the end of |f(gamma(t))| i found e^(2cost)/2

    And at the end of |(gamma'(t)| i founf 2.

    But i don't understand how can i integrate e^(2cost)

    Thank you

    Richard
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  2. #2
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    Use the ML inequality right? If |f(z)|\leq M on the contour \gamma and the length of the contour is L then:

    \left|\int_{\gamma} f(z)dz\right| \leq ML

    On the unit circle z=e^{it}=x+iy, |e^z|=|e^x| |e^{iy}|\leq e^1. Therefore \left|\frac{e^z}{z}\right|\leq e.
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  3. #3
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    Yes thank you very much.

    Here z=2e^(it) so is true that |e^z|<=e^2 ??

    So M=e^2 but why L should be pi??
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  4. #4
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    I need help plz..
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  5. #5
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    Quote Originally Posted by rickgoz View Post
    Yes thank you very much.

    Here z=2e^(it) so is true that |e^z|<=e^2 ??

    So M=e^2 but why L should be pi??
    If z=2e^{it} then |e^z|\leq e^2 but can you show that?

    L is still the length of the contour which is 2\pi r but I believe in the original problem, since 2<e, then it was true for \pi e^2 right?
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  6. #6
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    Okay thank you very much. It's fine!

    Last question: do you know how can I evaluate

    |14 + z + z^2| because I don't know if I can use the triangle inequality here..

    Thanks for your help

    Richard
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  7. #7
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    You can use the generalized version:

    |z_1+z_2+\cdots+z_n|\leq |z_1|+|z_2|+\cdots+|z_n|

    so |14+z+z^2|\leq 14+|z|+|z^2|
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