Consider $f$ nonconstant holomorphic on $\overline{B}(z_0,r)$, with $|f(z)|=R$ on $\partial B(z_0,r)$. Prove: $B(0,R)\subseteq f(B(z_0,r))$.

Question

Consider $f$ a nonconstant holomorphic function on $\overline{B}(z_0,r)$. If $|f(z)|=R$ on $\partial B(z_0,r)$ $(R>0)$, show that:

1. $f$ has a zero in $B(z_0,r)$.

2. $B(0,R)\subseteq f(B(z_0,r))$.

My attempt:

1. Suppose that $f$ has no zeroes in $B(z_0,r)$ then $1/f$ is holomorphic $\overline{B}(z_0,r)$. By the maximum modulus principle, we find $$ \left|\frac{1}{f(z)} \right|\le \max_{z\in\partial B(z_0,r)}\left|\frac{1}{f(z)} \right|=\frac 1R, \forall z\in \overline{B}(z_0,r).$$ From here, it follows that $|f|$ reaches its minimum (which is equal to $R$) on the boundary. However, $|f|$ also reaches its maximum on the boundary (by MMP), and it is also equal to $R$. Therefore $|f|=R$ on the compact disk, implying that $f$ is constant too - a contradiction.

2. In particular, we notice that $f$ is open on $B(z_0,r)$. This means that $$ (\forall z\in B(z_0,r))(\forall \varepsilon>0)(\exists \delta>0)(B(f(z),\delta)\subseteq f(B(z,\varepsilon))).$$ Choose $z=z_0$ and $\varepsilon=r$. From 1. we know that $f$ is a zero in this open disk, so it suffices to show that $f(z_0)=0$ and that the appropriate radius $\delta=R$. This is a tricky step. I was thinking Rouché, but I can't find good functions to apply it. Any hints?

Thanks.

Answer 1

Your proof of (1) is fine. For (2) assume for simplicity that $z_0 = 0$ and $R=1$.

For an $a \in B(0, 1)$ consider the function $g = T_a \circ f$, where $$ T_a(w) = \frac{w-a}{1-\overline a z} $$ is an automorphism of the unit disk. Part (1) applied to the function $g$ shows that $g(z_0) = 0$ for some $z_0 \in B(0, 1)$, and then $f(z_0) = a$, i.e. $a \in f(B(0,1 ))$.