Re: Help!!

[Arshad Momen]

The "policital" problem affects me as well.

For the first problem all the roots are double so use the formula :

\int F(z) /(z-a)^{n+1} dz = 2\pi i /(n!) F^{n}(a)

The second problem has all its roots as simple. Why do you claim to have discovered non-simple poles?

Post these msgs in the group instead to me only . I am not running a personal service.

On Wed, Apr 10, 2013 at 4:41 PM, pretty maria <prettym...@hotmail.com> wrote:

i m sorry for bothering you with these emails,but this is the only way i could get help while these political issues are in action..

  Da way i m  MARIA

     

 

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Date: Mon, 8 Apr 2013 16:21:21 +0600
Subject: RE: Help!!
From: arshad...@gmail.com
To: prettym...@hotmail.com

Get rid of that stupid SIG while posting in the group. We are not a chatroom.

On Apr 8, 2013 4:20 PM, "Arshad Momen" <amo...@univdhaka.edu> wrote:

Your function has double poles. The simple formula used in class won't work. You have to use Taylor expansion of the denominator near the singularities to get your residue. You can also use the second formula that one derives from the Cauchy formula by differentiating both sides with respect to a. Its in the book and you have created your own problem.

On Apr 8, 2013 4:15 PM, "pretty maria" <prettym...@hotmail.com> wrote:

here is the pdf..

  Da way i m  MARIA

     

 

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Date: Mon, 8 Apr 2013 15:46:18 +0600
Subject: Re: Help!!
From: amo...@univdhaka.edu
To: prettym...@hotmail.com

I can't read ur typed text. Send me a PDF of your formulas with proper fonts. My net is down and I am accessing mail on the dinky phone. Not the best Device to decipher math formulas.

On Apr 8, 2013 3:40 PM, "pretty maria" <prettym...@hotmail.com> wrote:

Assalam-o-alaikum sir...i have been doing a few problems on the cauchy residue theorem and got confused..
how about a problem such as  1/ (4+z^2)^2 ??  i tried doing this through knowledge from last class's lecture..
with z1=2e^ipi/2
         z2=2e^ipi
         z3=2e^3pi/2
         z4=2e^2pi
 
g'(z)= 2(4+z^2) * 2z
 
but as is find the res as z-> 2e^ipi/2 i get g'(z)=0 making it infinite...??..

  Da way i m  MARIA