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I have found the three points of singularity of the functions $$f(z)=\frac{z}{z^3+1}$$ as $z_1=e^{iπ/3}$, $z_2=e^{iπ}$, $z_3=e^{i5π/3}$

But what contour should I take to have the real axis from $0$ to $\infty$ as a part of my contour. I tried to take the quarter circle in the 1st quadrant with $R$ tending to $\infty$ which has one point $z_1$ inside my contour. But then when I split it into parts, I am not able to resolve further... Please tell whether I am approaching correctly or not.

санкет мхаске
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    Take instead the wedge of angle $2\pi/3$. Motivation; $(x\cdot e^{2\pi i/3})^3=x^3$ so your integrand is hardly affected – FShrike Sep 10 '23 at 07:22

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Well you can either try to make a substitution to collapse everything into 1 pole and use a keyhole contour, or you can try a pizza slice contour so something like this

enter image description here

since afaik it's likely that the ray coming back at $\theta = 2\pi /3$ in is going to be well behaved for your function and end up as a multiple of your integral.

The outer arc should just go to 0 since your numerator grows slower than denominator.

Max0815
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