# Finding integral using Residue Thm

Trying to Find $$\displaystyle \int_{\gamma}\frac{dz}{(z-3)(z^n-1)}$$ where $$\gamma$$ is the circle at origin with radius 2.

So I know $$\displaystyle \int_{\gamma}\frac{dz}{(z-3)(z^n-1)}=2\pi i (res(f,3)+\sum_{p \text{ root of unity} } res(f,p))$$

I did find that first part $$res(f,3)$$ is $$\frac{1}{3^n-1}$$ but I have no idea on how to approach the second part. Can someone help me on this with details?

Thanks!

• Each pole at a root of unity is a simple pole. Use the limit formula + L'Hopital Dec 4, 2022 at 11:42

The Residue Theorem tells you that the integral of $$f$$ along a simple closed curve equals $$2\pi i$$ times the sum of residues at points inside that curve. So you have to get rid of $$\text{Res}(f,3)$$, since $$3$$ lies outside the circle $$\gamma$$.
On the other hand, finding the residues at the roots of unity would be very tedious. Instead, consider $$\frac{1}{z^2}f\left(\frac{1}{z}\right)=\frac{z^{n-1}}{(1-3z)(1-z^n)},$$ and show that $$\text{Res}(f,\infty)=0$$. Since the sum of all residues is zero, it follows that $$\sum_{p\text{ roots of unity}}\text{Res}(f,p)=-\text{Res}(f,3).$$ Therefore, the integral becomes $$\int_\gamma \frac{dz}{(z-3)(z^n-1)}=-2\pi i\text{Res}(f,3).$$
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