For $f$ analytic on $\mathbb{D}$, if $f(\mathbb{D}) \subset \{ z \in \mathbb{C} : \Re(z) < 0\}$ and $f(0) = -1$, prove an inequality of $|f(z)|$.

Question

Old complex analysis qual question, here. The unit disk $\{z \in \mathbb{C} : |z| < 1\}$ is denoted $\mathbb{D}$:

For $f$ non-constant and analytic on $\mathbb{D}$, if $f(\mathbb{D}) \subset \{ z \in \mathbb{C} : \Re(z) < 0\}$ and $f(0) = -1$, prove that $$\frac{1 - |z|}{1 + |z|} \le |f(z)| \le \frac{1 + |z|}{1 - |z|}$$

I feel like I'm not seeing something obvious here. I can't compose this with some function from the RMT, since the image is not necessarily simply connected... Any hints?

Answer 1

Let $g(z)=\frac {f(z)+1} {f(z)-1}$. Then $g$ maps $\mathbb D$ into itself and vanishes at $0$. By Schwartz Lemma $|g(z)|\leq |z|$. This gives $|f(z)|-1 \leq |f(z)+1| =|g(z)| |f(z)-1| \leq|z|(|f(z)|+1)$. From this you get the right hand inequality. Now $\frac 1 f$ satisfies the hypothesis too and applying the right hand inequality to this function gives theleft hand inequality.