# Thread: Using residue calculus when poles are not exactly known

1. ## 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.

$\mathop\oint\limits_{|z|=2}\frac{1}{1.3\sin(1.7z)+ 0.6\sin(4z)}dz$

2. 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.

3. Originally Posted by Opalg
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?

4. Originally Posted by Opalg
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.