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Math Help - Using residue calculus when poles are not exactly known

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    Using residue calculus when poles are not exactly known

    Can the poles of this integrand be calculated exactly? I don't think so but not sure. How might one still use residues to solve (approximately) the integral even if the poles are only approximated? I posted a relevant plot which I'll leave unexplained just for fun for those interested.

    \mathop\oint\limits_{|z|=2}\frac{1}{1.3\sin(1.7z)+  0.6\sin(4z)}dz
    Attached Thumbnails Attached Thumbnails Using residue calculus when poles are not exactly known-approx-residued.jpg  
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    For the function 1/f(z), if there is a simple pole at z_0 then the residue there is 1/f'(z_0). So if you can get a good approximation to the location of the poles then you can also get a reasonable approximation to the values of the residues.
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    Quote Originally Posted by Opalg View Post
    For the function 1/f(z), if there is a simple pole at z_0 then the residue there is 1/f'(z_0). So if you can get a good approximation to the location of the poles then you can also get a reasonable approximation to the values of the residues.
    Can you provide the error terms?
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    Quote Originally Posted by Opalg View Post
    For the function 1/f(z), if there is a simple pole at z_0 then the residue there is 1/f'(z_0).
    Oh yea, I forgot about that. So let's say the upper right pole is about 1+0.25i. Then using Mathematica's \text{FindRoot}, I get for the residue there (last line):

    Code:
    In[7]:= f[z_] = 1.3*Sin[1.7*z] + 0.6*Sin[4*z]; 
    urroot = FindRoot[f[z] == 0, {z, 1 + 0.25*I}]
    r1 = 1/D[f[z], z] /. urroot
    
    Out[8]= {z -> 1.2794588067848602 + 
           0.3743082054455279*I}
    
    Out[9]= 0.05685562855624362 - 0.2787028735656702*I
    Suppose could do that 4 more times and get a good approximation to the integral via residues. Just news to me, that's all.
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