I'm studying "Meromorphic Functions" by W. K. Hayman and I'm stuck with the proof for theorem 1.7.1 (p.18):
If $f$ is regular for $|z|\leqslant R$ and $$M(r,f)=\max_{|z|=r}|f(z)|,$$ then $$T(r,f)\leqslant\log^+M(r,f)\leqslant\frac{R+r}{R-r}R(R,f)\quad(0\leqslant r<R).$$
It is written with Nevanlinna's notation, which is $$\log^+ x=\left\{\begin{array}{cll} 0 & \text{if} & 0\leqslant x<1;\\ \log x & \text{if} & x\geqslant1. \end{array}\right.$$ $$m(R,f)=\frac{1}{2\pi}\int_0^{2\pi}\log^+|f(Re^{i\phi})|\,d\phi;$$ $$N(R,f)=\sum_{\nu=1}^N\log\left|\frac{R}{b_\nu}\right|;$$ where $b_\nu$, $\nu=1,\ldots,n$ are the poles of $f$ in $|z|<R$ (counting their multiplicities); and $$T(R,f)=m(R,f)+N(R,f).$$ My problem is here:
Suppose that $M(r,f)>1$ and chose $z_0=re^{i\theta}$ so that $|f(z_0)|=M(r,f)$. Since $f(z)$ has no poles in $|z|<R$, Theorem 1.1 [Poisson-Jensen formula] yields $$\log^+M(r,f)=\log|f(z_0)|\leqslant\frac{1}{2\pi}\int_0^{2\pi}\log|f(Re^{i\phi})|\frac{(R^2-r^2)\,d\phi}{R^2-2Rr\cos(\theta-\phi)+r^2}\leqslant\cdots$$
Well, if I use the Poisson-Jensen formula, I will get, as there is no poles,
$$\log|f(z_0)|=\frac{1}{2\pi}\int_0^{2\pi}\log|f(Re^{i\phi})|\frac{(R^2-r^2)\,d\phi}{R^2-2Rr\cos(\theta-\phi)+r^2}+\sum_{\mu=1}^M\log\left|\frac{R(z_0-a_\mu)}{R^2-\bar a_\mu z_0}\right|,$$ where now $a_\mu$, $\mu=1,\ldots,M$ are the zeros of $f$ in $|z|<R$ (again, counting their multiplicities).
For the inequality to be true we need the following condition to be satisfied, for every zero $a_\mu$, $$\left|\frac{R(z_0-a_\mu)}{R^2-\bar a_\mu z_0}\right|\leqslant1.$$ But I can't see why this is true. Any hint will be welcome.
Divide numerator and denominator by $R^2$:
$$\frac{R(z_0-a_{\mu})}{R^2-\overline{a}_{\mu}z_0}= \frac{\dfrac{z_0}{R}-\dfrac{a_{\mu}}{R}}{1-\dfrac{\overline{a}_{\mu}}{R}\frac{z_0}{R}} $$
Now, realize it's just an automorphism of the unit disk applied to the point $z_0/R$, which has modulus $r/R<1$. Its image by that automorphism will have also modulus less than one.