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Math Help - modulus of a line integral bounded from above

  1. #1
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    modulus of a line integral bounded from above

    \

    did i do this one correctly?
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  2. #2
    Member mabruka's Avatar
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    Well i would use something like:

     \left| \int_{\delta R} \frac{u(z)}{(z-z_0)^2} dz \right| \leq  CR\int_{0}^{2\pi}    \left|   \frac{1}{(Re^{it}-z_0)^2} dt \right| \leq \frac{2\pi CR}{||R|-|z_0||^2}


    the last part goes to 0 as R grows


    What do u think?


    I think there is something wrong with your attempt, your bound must not depend on t.
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  3. #3
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    Quote Originally Posted by mabruka View Post
    Well i would use something like:

     \left| \int_{\delta R} \frac{u(z)}{(z-z_0)^2} dz \right| \leq  CR\int_{0}^{2\pi}    \left|   \frac{1}{(Re^{it}-z_0)^2} dt \right| \leq \frac{2\pi CR}{||R|-|z_0||^2}


    the last part goes to 0 as R grows


    What do u think?


    I think there is something wrong with your attempt, your bound must not depend on t.

    I agree. Can you explain, though, why (Re^{it}-z_0)^2 = ||R|-|z_0||^2 is true?
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  4. #4
    Member mabruka's Avatar
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    Well i thouht using triangle inequality and then just squaring it up:




    ||Re^{it}|-|z_0||\leq | Re^{it}- z_0| \Rightarrow \frac{1}{| Re^{it}- z_0| } \leq \frac{1}{||Re^{it}|-|z_0||}=\frac{1}{||R|-|z_0||}

    where obviously |Re^{it}|=|R|=R


    What do you think?
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  5. #5
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    Ah yes, you're absolutely right. Thank you so much, this makes perfect sense to me now.
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