I am trying to prove that
$$ \int_C \frac{P(z)}{Q(z)} dz = 0 $$
If polynomial order of $Q$ is 2 or more than that of $P$, using the theorem stating that if a function has a finite number of singular points all interior to a contour $C$, then
$$ \int_C f(z) dz = 2\pi i\text{Res}\bigg[\frac{1}{z^2} f(\frac{1}{z})\bigg] $$
I received the hint that, under the conditions described above, the rational function in the integrand can be written as a McLauran Series with no negative powers of z. This would imply that the residue is zero...
My problem is, I can't seem to wrap my head around how the rational function given to me can be written into the form of a power series with only positive exponents...
So, as the title says, When can a rational function be represented as a power series? I'm not looking for a full proof, but a few details would be nice, just so I feel more comfortable running with the hint.
The answer is that whenever you center the expansion at a point where the function is analytic, it can be represented as a Taylor series. In this specific case, we will see that the function in question is analytic at $z=0$ and therefore has a Maclaurin expansion.
To see why we can write $\frac{1}{z^2}\frac{P(\frac{1}{z})}{Q(\frac{1}{z})}$ as a Taylor series centered at $z=0$, consider distributing the $\frac{1}{z^2}$ onto the rational function as shown below:
$$ g(z)=\frac{1}{z^2}\frac{P(\frac{1}{z})}{Q(\frac{1}{z})} = \frac{a_0 + a_1\frac{1}{z} + \dots + a_n\frac{1}{z^n}}{b_0z^2 + b_1z + b_2 + \dots+b_m\frac{1}{z^{m-2}}} \times \frac{z^{m-2}}{z^{m-2}} = \frac{a_0z^{m-2} + a_1z^{m-3} + \dots + a_nz^{m-n-2}}{b_0z^m + b_1z^{m-1} + \dots+b_m} $$
Now, take note of the far right hand side of the above equation and consider what happens when $z=0$. Because we know $b_m$ is non zero, the denominator is non-zero at $z=0$. This means that $g(z)$ is analytic at $z=0$
We know then that if a function is analytic within some neighborhood of $z=0$ that it can be written as a Maclaurin power series:
$$ g(z) = \sum_{n=0}^\infty d_n z^n $$
Because there are no negative powers of $z$ in this expansion, its residue is zero at $z=0$.
Using the above information and the fact that $\frac{P(z)}{Q(z)}$ is analytic except for a finite number $n$ of singular points (the zeros of $Q$) contained in a large enough contour $C$, then:
$$ \int_C f(z)dz = 2\pi i \sum_{j=1}^n\underset{z = z_j}{\text{Res}}[f(x)] = 2\pi i { \underset{z = 0}{\text{Res}}\bigg[\frac{1}{z^2}f\bigg(\frac{1}{z}\bigg)\bigg]} = 0 $$