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Finding an upper bound for function over square contour Watch

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    The function is
     f(z)=\frac{cosec(\pi*z)}{z^2}
    and the questions is "by finding a suitable upper bound for |f(z)| on  C_N Where  C_N is the square with vertices at the points  (N+\frac{1}{2})(\pm1 \pm i) for N \geq 1 show that  \int f(z) \ dz \rightarrow 0 as  x \rightarrow \infty .

    My attempt:
    I said that if z is on the square then  | \frac{1}{z^2} | \leq \frac{4}{9}

as  |z| \geq \frac{3}{2}
    I then expressed  | \frac{1}{sin(\pi*z)} | in exponential form with z = x+iy and x=(N+1/2) to find the upper bound when z is on one of the vertical sides. I got  | \frac{1}{sin(pi*z)} | = \frac{\mathrm{2e}^{(-y*\pi)}}{1+ \mathrm{e}^{(2y*\pi)}} \leq \mathrm 2{e}^{-y\pi}.

    so  |f(z)| < \frac{8}{9} \mathrm{e}^{-y\pi} on the vertical sides of the square.

    Am I doing this right?
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    (Original post by Muffin.)
    The function is
     f(z)=\frac{cosec(\pi*z)}{z^2}
    and the questions is "by finding a suitable upper bound for |f(z)| on  C_N Where  C_N is the square with vertices at the points  (N+\frac{1}{2})(\pm1 \pm i) for N \geq 1 show that  \int f(z) \ dz \rightarrow 0 as  x \rightarrow \infty .

    My attempt:
    I said that if z is on the square then  | \frac{1}{z^2} | \leq \frac{4}{9}

as  |z| \geq \frac{3}{2}
    I then expressed  | \frac{1}{sin(\pi*z)} | in exponential form with z = x+iy and x=(N+1/2) to find the upper bound when z is on one of the vertical sides. I got  | \frac{1}{sin(pi*z)} | = \frac{\mathrm{2e}^{(-y*\pi)}}{1+ \mathrm{e}^{(2y*\pi)}} \leq \mathrm 2{e}^{-y\pi}.

    so  |f(z)| < \frac{8}{9} \mathrm{e}^{-y\pi} on the vertical sides of the square.

    Am I doing this right?

    I think you need to somehow involve  N in when you are estimating  | \frac{1}{z^2} | ? s.t. when  N \rightarrow \infty , you will have  \int\Limits_{C_N} \! f(z) \ dz \rightarrow 0

    btw did you learn latex in your free time or does your university teaches you?
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    (Original post by WKH)
    I think you need to somehow involve  N in when you are estimating  | \frac{1}{z^2} | ? s.t. when  N \rightarrow \infty , you will have  \int\Limits_{C_N} \! f(z) \ dz \rightarrow 0

    btw did you learn latex in your free time or does your university teaches you?
    Thanks for that :-)
    And yep I used the TSR wiki on LaTex lol
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    (Original post by Muffin.)
    Thanks for that :-)
    And yep I used the TSR wiki on LaTex lol
    How do you get on with it?

    I think if you consider  |z| is on the edge of the square then you can estimate  |\frac{1}{z^2}|
 
 
 
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