I'm studing complex analysis.I saw that
$R(z)$ is a rational function.Consider the function $R(1/z)$ which we can rewrite as a rational function $R_1(z)$,and set $R(\infty)=R_1(0).$
with the notation $$R(z)=\frac{a_0+a_1z+\ldots+a_nz^n}{b_0+b_1z+\ldots+b_mz^m}$$
we obtain $$R_1(z)=\frac{a_0z^n+a_1z^{n-1}+\ldots+a_n}{b_0z^m+b_1z^{m-1}+\ldots+b_m}z^{m-n}$$
After "we obtain" I am not sure what is happening . Could anyone show more detail about it?Thanks!
See what happens when you substitute in for $R(1/z)$. Given the definition of $R$, this will give us:
$$R(1/z)=\frac{a_0+a_1z^{-1}+\ldots+a_nz^{-n}}{b_0+b_1z^{-1}+\ldots+b_mz^{-m}}$$
We can rewrite the powers of $z$ in the numerator and denominator as follows. Note that all I have done here is re-express the exponents of $z$: $0 = n-n$, $-1 = (n-1) -n$, etc.
$$R(1/z)=\frac{a_0 z^{n-n} + a_1 z^{(n-1)-n} + \ldots + a_n z^{0-n}}{b_0 z^{m-m} +b_1 z^{(m-1)-m} + \ldots + b_m z^{0-m}}$$
And hence, we can pull some factors of $z$ out of the fraction:
$$R(1/z)=\frac{a_0 z^n + a_1 z^{n-1} + \ldots + a_n z^{0}}{b_0 z^{m} +b_1 z^{m-1} + \ldots + b_m z^{0}} \cdot \frac{z^{-n}}{z^{-m}}$$
Rewriting the fraction on the right as a simple power of $z$, we get:
$$z^{-n}/z^{-m} = z^m/z^n = z^{m-n}$$
Letting $R_1(z)$ be defined as $R(1/z)$, we finally get the result that "[they] obtain":
$$R_1(z) := R(1/z)=\frac{a_0z^n+a_1z^{n-1}+\ldots+a_n}{b_0z^m+b_1z^{m-1}+\ldots+b_m}z^{m-n}$$