= Solution
Put $\alpha=-\log|c|>0$. Since $f$ has no zeros and the <unit disc> is <simply connected>, it has a <holomorphic logarithm>. Thus $F=-\log f$ is <holomorphic> with $\operatorname{Re}F>0$ and $\operatorname{Re}F(0)=\alpha$. The <Cayley transform between the half-plane and disk> gives
$$
\phi(z)=\frac{F(z)-F(0)}{F(z)+\overline{F(0)}}\in\mathbb D,\qquad \phi(0)=0.
$$
The <Schwarz lemma> yields $|\phi'(0)|\le1$, so the <Carathéodory derivative bound for the right half-plane> gives $|F'(0)|\le2\alpha$. Since $f'=-F'f$, the sharp answer is
$$
\boxed{\max |f'(0)|=2|c|\log\frac1{|c|}.}
$$
Equality in the <Schwarz lemma> holds exactly when $\phi(z)=\omega z$, $|\omega|=1$. Solving for $F$ and exponentiating gives exactly the extremizers
$$
\boxed{f(z)=c\exp\left(-\frac{2\alpha\omega z}{1-\omega z}\right),\qquad |\omega|=1.}
$$
Their logarithmic real part is $\alpha\operatorname{Re}[(1+\omega z)/(1-\omega z)]>0$, so each is indeed a zero-free map into the <unit disc> with the prescribed value at zero.
For a fixed $w\in\mathbb D$, put $r=|w|$. Apply the <Harnack inequality on the unit disk> to the positive <harmonic function> $\operatorname{Re}F$ and then exponentiate its negative. The <zero-free Schwarz lemma> gives
$$
\boxed{|c|^{(1+r)/(1-r)}\le |f(w)|\le |c|^{(1-r)/(1+r)}.}
$$
For $w\ne0$ both endpoints are attained: take $\omega w=r$ for the lower endpoint and $\omega w=-r$ for the upper endpoint in the displayed extremizers. At zero both are $|c|$. If the question is interpreted as allowing $w$ to vary as well, the sharp <infimum> and <supremum> over all admissible pairs $(f,w)$ are respectively $0$ and $1$, approached as $r\uparrow1$; neither value is attained inside the <unit disc>. An individual function need not have these global extremal values: the constant function $f=c$ is a simple example.
Back to article page
