Question regarding asymptotics of complex rational functions Suppose $P\left(z\right)$
  and $Q\left(z\right)$
  are complex polynomials such that $\deg Q=m\geq l=\deg P$
  and wlog suppose the lead coefficient in both polynomials is $1$. I want to show that $\left|\frac{p\left(z\right)}{q\left(z\right)}\right|$
  (which is a function taking real values) is asymptotically equivalent to $\left|\frac{1}{z^{m-l}}\right|$
  in the sense that $$\left|\left|z^{m-l}\right|\left|\frac{p\left(z\right)}{q\left(z\right)}\right|\right|\longrightarrow1$$
  as $z\to\infty$
 . Obviously getting this result for real valued rational functions is extremely easy but I'm having some difficulty showing this for complex rational functions. I tried writing $P\left(z\right)=\sum_{i=0}^{l}a_{i}z^{i}$
  and $Q\left(z\right)=\sum_{i=0}^{m}b_{i}z^{i}$
  and then writing: $$\left|z^{m-l}\frac{P\left(z\right)}{Q\left(z\right)}\right|=\left|\frac{\frac{z^{m-l}}{z^{m}}P\left(z\right)}{\frac{1}{z^{m}}Q\left(z\right)}\right|=\left|\frac{\sum_{i=0}^{l}a_{i}z^{i-l}}{\sum_{i=0}^{m}b_{i}z^{i-m}}\right|$$
 From this it's very easy to get that: $$\frac{\sum_{i=0}^{l}\left|a_{i}z^{i-l}\right|}{\sum_{i=0}^{m}\left|b_{i}z^{i-m}\right|}\overset{z\to\infty}{\longrightarrow}\frac{a_{l}}{b_{m}}=\frac{1}{1}=1$$
But that's not what I want to show and I can't see if I can use this to get the required result.
EDIT - Additional question: Furthermore, under these assumptions I would like to say that $\sup_{\left|z\right|=R}\left|\frac{P\left(z\right)}{Q\left(z\right)}\right|$, acts like $\sup_{\left|z\right|=R}\left|\frac{1}{z^{m-l}}\right|=\frac{1}{R^{m-l}}$ as $R\to\infty$ how would I go about doing that?
 A: The same approach for real polynomials works. We proceed by induction on $m$. Clearly the result holds if $m=0$. We have
$$\left|\frac{z^m}{Q(z)}\right|-\left|\frac{a_{l-1}z^{m-1}+\cdots+a_1z^{m-l-1}}{Q(z)}\right|\le \left|\frac{z^{m-l}P(z)}{Q(z)}\right|\le \left|\frac{z^m}{Q(z)}\right|+\left|\frac{a_{l-1}z^{m-1}+\cdots+a_1z^{m-l-1}}{Q(z)}\right|$$
so it suffice to show that
$$\left|\frac{z^m}{Q(z)}\right|\to 1 \;\;\text{and}\;\; \left|\frac{a_{l-1}z^{m-1}+\cdots+a_1z^{m-l-1}}{Q(z)}\right|\to 0.$$
For the first limit, note that this is equivalent to 
$$\left|\frac{Q(z)}{z^m}\right|\to 1$$
which follows from $|z^m/z^m|\to 1$ (which is trivial) and 
$$\left|\frac{b_{m-1}z^{m-1}+\cdots+a_1}{z^m}\right|\to 0$$
which is a special case of the second. For the second, note that this is equivalent to 
$$\left|\frac{Q(z)}{a_{l-1}z^{m-1}+\cdots+a_1z^{m-l-1}}\right|\to \infty$$
which follows from 
$$\left|\frac{z^m}{a_{l-1}z^{m-1}+\cdots+a_1z^{m-l-1}}\right|\ge \frac{|z|^m}{|a_{l-1}||z|^{m-1}+\cdots+|a_1||z|^{m-l-1}}\to \infty$$
and the fact that, by the inductive hypothesis, we have
$$\left|\frac{b_{m-1}z^{m-1}+\cdots+b_1}{a_{l-1}z^{m-1}+\cdots+a_1z^{m-l-1}}\right|\to \left|\frac{b_{m-1}}{a_{l-1}}\right|.$$
