Suppose $f: \mathbb{D} \to \mathbb{D}$ is a holomorphic function and $f(0)=0$. The function $f$ has a total of five zeros (counting multiplicities) in the closed half-disc $\overline{\frac{1}{2}\mathbb{D}} = \{z \in \mathbb{C}: |z| \leq \frac{1}{2}\}$. How large can $|f'(0)|$ be?

Schwarz Lemma $f: \mathbb{D} \to \mathbb{C}$ is analytic with $f(0) = 0$ and $|f(z)| \leq 1$ on $\mathbb{D}$, then $|f(z)| \leq |z|$ for all $z \in \mathbb{D}$ and $|f'(0)| \leq 1$

It would be almost indisputable that Schwarz lemma will be of use here. Now, given that $f$ has five zeros in the domain $\overline{\frac{1}{2}\mathbb{D}}$, it's no question that Rouche's theorem will likely also be of use here.

Since $f$ maps $\mathbb{D}$ to itself, then implicitly we also have $|f(z)| \leq 1$ for $z$ on $\mathbb{D}$. Then $f$ satisfies Schwarz's lemma, and so an upper bound for $|f'(0)|$ is $1$. Perhaps this bound can be improved.

But I have not used the fact that $f$ has five zeros in $\overline{\frac{1}{2}\mathbb{D}}$. Again, I know this must be Rouche's theorem, but I'm not sure how to apply it.

  • 1
    $\begingroup$ The answer is $1/16$ - hint: the problem shows that $f=zB(z)e^{g(z)}$ where $B$ is a Blaschke product with at least four zeroes in the half disc and $\Re g\le 0$; when you take the derivative at $0$ only $B(0)e^{g(0)}$ counts etc $\endgroup$
    – Conrad
    Commented May 14 at 14:14

1 Answer 1


Let $a_1, \ldots, a_4$ denote the four zeros of $f$ with $0 < |a_k| \le 1/2$. Then $$ f(z) = h(z) \prod_{k=1}^n \frac{z-a_k}{1-\overline{a_k} z} $$ with $h: \Bbb D \to \Bbb D$ and $h(0) = 0$. It follows that $$ f'(0) = h'(0) \prod_{k=1}^n (-a_k) \, . $$ We have $|h'(0)| \le 1$ by the Schwarz lemma, so that $$ |f'(0)| \le \prod_{k=1}^n |a_k| \le \frac{1}{2^4} = \frac{1}{16} \, . $$ The bound is sharp, equality holds if and only if $$ f(z) = c z \prod_{k=1}^n \frac{z-a_k}{1-\overline{a_k} z} $$ with complex numbers $c, a_1, \ldots, a_4$ satisfying $|c|=1$ and $|a_1|=|a_2|=|a_3|=|a_4| = 1/2$.

  • $\begingroup$ I actually have never seen the first equality. How does one arrive there? Elegant answer though.... $\endgroup$ Commented May 15 at 7:36
  • $\begingroup$ @HyperbolicPDEfriend: For $|a| < 1$ is $T_a(z) = \frac{z-a}{1-\bar a z}$ an automorphism of the unit disk (sometimes called “Blaschke factor”). If $f$ maps the unit disk into itself with $f(a) = 0$ then $f(z)/T_a(z)$ has a removable singularity and still maps the unit disk into itself (by the maximum modulus principle). Here $h(z) = \frac{f(z)}{z\prod_{k=1}^4 T_{a_k}}(z)$ has removable singularities and maps the unit disk into itself. The product in the denominator is sometimes called “finite Blaschke product.” $\endgroup$
    – Martin R
    Commented May 15 at 9:39
  • $\begingroup$ Thank you so much! $\endgroup$ Commented May 15 at 9:51
  • $\begingroup$ Can you justify $f'(0)= h'(0)\displaystyle\prod_{k=1}^n (-a_k)$? It's not clear to me how this is the case. The way this is written makes it look like the derivative is multiplicative, which I'm sure is merely a coincidence $\endgroup$ Commented May 25 at 14:05
  • $\begingroup$ @GrigorHakobyan: It is just the product rule applied to $f(z) = h(z)T_1(z) \cdots T_n(z)$. Since $h(0) = 0$, the result is $f'(0) = h'(0)T_1(0) \cdots T_n(0)$. $\endgroup$
    – Martin R
    Commented May 25 at 14:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .