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Finding an upper bound for function over square contour

The function is
f(z)=cosec(πz)z2[br] f(z)=\frac{cosec(\pi*z)}{z^2}[br]
and the questions is "by finding a suitable upper bound for |f(z)| on CN C_N Where CN C_N is the square with vertices at the points (N+12)(±1±i)forN1 (N+\frac{1}{2})(\pm1 \pm i) for N \geq 1 show that f(z) dz0 \int f(z) \ dz \rightarrow 0 as x x \rightarrow \infty .

My attempt:
I said that if z is on the square then 1z249[br] | \frac{1}{z^2} | \leq \frac{4}{9}[br] as z32 |z| \geq \frac{3}{2}
I then expressed 1sin(πz) | \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 1sin(piz)=2e(yπ)1+e(2yπ)2eyπ | \frac{1}{sin(pi*z)} | = \frac{\mathrm{2e}^{(-y*\pi)}}{1+ \mathrm{e}^{(2y*\pi)}} \leq \mathrm 2{e}^{-y\pi}.

so f(z)<89eyπ |f(z)| < \frac{8}{9} \mathrm{e}^{-y\pi} on the vertical sides of the square.

Am I doing this right? :s-smilie:
Reply 1
Original post by Muffin.
The function is
f(z)=cosec(πz)z2[br] f(z)=\frac{cosec(\pi*z)}{z^2}[br]
and the questions is "by finding a suitable upper bound for |f(z)| on CN C_N Where CN C_N is the square with vertices at the points (N+12)(±1±i)forN1 (N+\frac{1}{2})(\pm1 \pm i) for N \geq 1 show that f(z) dz0 \int f(z) \ dz \rightarrow 0 as x x \rightarrow \infty .

My attempt:
I said that if z is on the square then 1z249[br] | \frac{1}{z^2} | \leq \frac{4}{9}[br] as z32 |z| \geq \frac{3}{2}
I then expressed 1sin(πz) | \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 1sin(piz)=2e(yπ)1+e(2yπ)2eyπ | \frac{1}{sin(pi*z)} | = \frac{\mathrm{2e}^{(-y*\pi)}}{1+ \mathrm{e}^{(2y*\pi)}} \leq \mathrm 2{e}^{-y\pi}.

so f(z)<89eyπ |f(z)| < \frac{8}{9} \mathrm{e}^{-y\pi} on the vertical sides of the square.

Am I doing this right? :s-smilie:



I think you need to somehow involve N N in when you are estimating 1z2 | \frac{1}{z^2} | ? s.t. when N N \rightarrow \infty , you will have
Unparseable latex formula:

\int\Limits_{C_N} \! f(z) \ dz \rightarrow 0



btw did you learn latex in your free time or does your university teaches you?
(edited 11 years ago)
Reply 2
Original post by WKH
I think you need to somehow involve N N in when you are estimating 1z2 | \frac{1}{z^2} | ? s.t. when N N \rightarrow \infty , you will have
Unparseable latex formula:

\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
Reply 3
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 |z| is on the edge of the square then you can estimate 1z2 |\frac{1}{z^2}|

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