$$ \int_0^{2 \pi} \frac{\rho e^{i \theta} d \theta}{\left(\rho e^{i \theta}-z\right)^{h+2}}=\frac{-i}{h+1}\left[\frac{1}{\left(\rho e^{i \theta}-z\right)^{h+1}}\right]_0^{2 \pi}=0 $$ $\therefore$ We can subtract $\log M(\rho)$ from $\log |f|$ If $\rho>2|z|$, if follows that the last term on the R.H.S. of (9) has a modulus at most equal to $$ \underline{(h+1)} 2^{h+3} \rho^{-h-1} \frac{1}{2 \pi} \int_0^{2 \pi} \log \frac{M(\rho)}{\left|f\left(\rho e^{i \theta}\right)\right|} d \theta \text { for } \log \left(\frac{M(\rho)}{\left|f\left(\rho e^{i \theta}\right)\right|}\right)>0 . $$
how to arrive the second step in this proof .?