Let $n \in \mathbb{N}_0$ and set $p(z) = z^n + a_1 z^{n-1} + \cdots$ and $q(z) = z^{n+1} + b_1 z^{n} + \cdots$ to be two monic complex polynomials with no common zeros. I want to prove that $$\frac{1}{2\pi i} \int_{\partial D_{R}(0)} \frac{p(z)}{q(z)}dz = 1,$$ for $R$ large enough. I've found the solution with the argument principle, but I started thinking: is there a way where you do not use the argument principle? I considered using Cauchy's theorem. You can use a transformation on both polynomials in order to get a constant term with value $+1$ and when $z= 0$ is a root of $q(z)$, the result easily follows. On the other hand, if $q(0) = a + bi$, then you can use the transformation $z \mapsto z - (a+bi)$ so that $q(0) =0$. The problem is, I don't know what to do next, or even know if this is the right lane of thought. Does anyone have an idea for this one?
Kind regards.
Hint.
You have $\frac{p(z)}{q(z)} = \frac{1}{z}+ \frac{1}{z}\epsilon(\frac{1}{z})$ where $\epsilon(Z) \to 0$ as $Z \to 0$. The result then follows from $$\int_{\partial D_R(0)} \frac{dz}{z} = 2 i \pi$$