Consider this function for $0 < a < b$:
$$f_{(z)} = \frac{z^4}{z^2(z-\frac{a}{b})(z-\frac{b}{a})}$$
This function has a pole of order $2$ at $z=0$, a pole of order 1 at $z=\frac{a}{b}$, but what about the pole at $z=\frac{b}{a}$? Why isn't it mentioned in the book?
Here's the original question:
Evaluate
$$I = \int_0^{2\pi} \frac{\cos 2\theta}{a^2 + b^2 -2ab \cos \theta} d\theta$$
Then by using subsitution of $z = e^{i\theta}$ we arrive at the form above.
Is the reason they are ignoring the $z=\frac{b}{a}$ term because it is greater than 1, so it lies outside the circle, where residue theorem doesn't work?
Notice that you can interchange $a$ and $b$ in the integral and the integrand remains unchanged. So without loss of generality, we can choose $b>a$. Then the substitution $z=\mathrm{e}^{\mathrm{i}\theta}$ parameterises the unit circle, which will contain the poles at $z=0$ and $z=a/b$, but not the pole at $z=b/a$.