Originally Posted by

**Opalg** For the function $\displaystyle 1/f(z)$, if there is a simple pole at $\displaystyle z_0$ then the residue there is $\displaystyle 1/f'(z_0)$.

Oh yea, I forgot about that. So let's say the upper right pole is about $\displaystyle 1+0.25i$. Then using Mathematica's $\displaystyle \text{FindRoot}$, 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.