Singularities of quotient of polynomials where the degree of the denominator $\ge$ the degree of the numerator $+2$. Let the degrees of the polynomials 
$$P(z)=a_0+a_1 z+a_2 z^2+\cdots +a_n z^n \; (a_n \neq 0)$$
and 
$$Q(z)=b_0+b_1 z+b_2 z^2+\cdots +b_m z^m \; (b_m\neq 0)$$
be such that $m \ge n+2.$ Show that if all of the zeros of $Q(z)$ are interior to a simple closed ontour $C$, then $\int_C \frac{P(z)}{Q(z)}dz=0.$
The solution to this problem states the following:
$$\frac{1}{z^2} \cdot \frac{P(1/z)}{Q(1/z)}=\frac{a_0 z^{m-2}+a_1 z^{m-3}+a_2 z^{m-4}+\cdots +z_n z^{m-n-2}}{b_0 z^m+b_1 z^{m-1}+b_2 z^{m-2}+\cdots +b_m} \; (z\neq 0).$$
Observe that the numerator here is in fact a polynomial since $m-n-2\ge 0.$ Also, since $b_m \neq 0$, the quotient of these polynomials is represented by a series of the form $d_0 + d_1 z+d_2 z^2+\cdots.$ That is, 
$$\frac{1}{z^2}\cdot \frac{P(1/z)}{Q(1/z)}=d_0 + d_1 z+d_2 z^2+\cdots \; (0\lt |z|\lt R_2);$$ has residue $0$ at $z=0$.
I have trouble understanding the bolded part, that is, why does $b_m\neq 0$ mean that the quotient is represented by a power series without any fractional powers of $z$?
 A: Since $b_m\neq 0$, the fraction $$\frac{a_0 z^{m-2}+a_1 z^{m-3}+a_2 z^{m-4}+\cdots +z_n z^{m-n-2}}{b_0 z^m+b_1 z^{m-1}+b_2 z^{m-2}+\cdots +b_m}$$ is defined and continuous at $z=0$ (the denominator does not vanish).  This means that the singularity of $$g(z)=\frac{1}{z^2} \cdot \frac{P(1/z)}{Q(1/z)}$$ at $z=0$ is removable, so $g(z)$ can be extended to be analytic at $0$.  Thus $g(z)$ has a Taylor series expansion in some disk around $0$.
A: The denominator polynomial $\Delta(z)=b_0 z^m+b_1 z^{m-1}+b_2 z^{m-2}+\cdots +b_m$
is continuous at the origin and non-zero there: $\Delta(0)=b_m\not=0.$
Of course $\Delta(z)$ is entire, so combining the observations we see that $\Delta(z)$ is analytic and non-zero in a disc $D(0,\epsilon).$
The numerator polynomial $N(z)$ is obviously analytic in the same disc $D(0,\epsilon).$ So in this disc the quotient of these polynomials is an analytic function and hence possesses a Taylor Series $d_0 + d_1 z+d_2 z^2+\cdots,$ valid in any disc with radius $\rho: 0<\rho\leqslant\epsilon$, by Taylor's Theorem.
This already answers your question about the bolded text.
Carrying on to the identity
$$\frac{1}{z^2} \cdot \frac{P(1/z)}{Q(1/z)}=\frac{a_0 z^{m-2}+a_1 z^{m-3}+a_2 z^{m-4}+\cdots +z_n z^{m-n-2}}{b_0 z^m+b_1 z^{m-1}+b_2 z^{m-2}+\cdots +b_m} \; (z\neq 0),$$
we note that the left-hand side is not defined if $z\not=0$ is such that $Q(1/z)=0.$
But we can choose $\rho$ very small so that such $z$ are not a worry. So really we should write
$$\frac{1}{z^2} \cdot \frac{P(1/z)}{Q(1/z)}=d_0 + d_1 z+d_2 z^2+\cdots \; (0<|z|<\rho).$$
Last of all we appeal to the Uniqueness Theorem: whenever a series of Laurent form $\sum d_j z^j$ converges throughout a given annulus, it is the Laurent series.
