So I take $R>0$ and I consider the contour $\Gamma_R$ generated by the half circle and [-R, R] in the upper half plane $\{Im(z) \geq 0\}$. Suppose $P \in \mathbb{C}[z]$ a polynomial of degree greater than 1 and without roots, by Cauchy's theorem, I have
$$\int_{\Gamma_R} \frac{dz}{P(z)\overline{P(\bar{z})}} = 0, \quad \forall R > 0$$
So $$0 = \int_{C_R}\frac{dz}{P(z)\overline{P(\bar{z})}} + \int_{I_R}\frac{dz}{P(z)\overline{P(\bar{z})}} = \int_{C_R}\frac{dz}{P(z)\overline{P(\bar{z})}} + \int_{I_R}\frac{dz}{|P(z)|^2} = \varphi(R) + \psi(R).$$
As deg(P) > 1, there exists $R_0 > 0$ and $C > 0$ such that $|z| > R_0 \Rightarrow |P(z)| \geq C|z|.$
How can I conclude from this that $$|\varphi(R)| \leq \int_{0}^{\pi}\frac{Rd\theta}{C^2 R^2} \leq \frac{\pi}{C^2 R}, \quad R > R_0?$$
All feedback is welcomed.